permutation and combination

P a g e | 1 Permutation and Combinations: Art of Counting

Combinatorics

The Art of Counting Best way to learn permutation and

Combination.

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Contents

Contents Page Numbers

Definition 2-4

Fundamental Principle of Counting 4-6

Multiplication Principal 6-10

Addition Principle 10-13

Bijection Principle 13-14

Inclusion-Exclusion Principle 14-16

Factorials 15-16

Permutations 16-19

Circular Permutations 19-33

Permutations involving Identical Objects 33-40

Exponent of Prime p in n! 40-43

Assignments (Subjective) 43-45

Assignment level II 45-49

Assignments (Objective) 49-74



Combinatorics :The Art of Counting

Introduction:.Combinatorics deals mainly with questions focused on counting. Questions like “In how many ways is it possible to...?” or “What is the possible number of …?”

Here are a few examples –

1. Bar Codes

You must have observed these stamped on almost everything you buy!

How many such different barcodes are possible?

How could the barcode be designed or modified so that there can be at least 1000000 different codes?

2. Vehicle Registration Numbers

How many vehicles can be registered, given the format of the registration number on the left?

10000? 1000000? 99999999? What do you feel it to be?

3. p@sSwoRd$

How much time would it require (worst case) to guess (or hack) a password or a secret code, which is 9 character long and can contain alphabets, numbers and some special characters like @#$%&*?

Well, one thing it will surely depend upon is – How many such different codes would be possible?



4. Smartphone Lock Screens

This looks familiar, doesn’t it?

Suppose you forget the code to unlock your screen, and you start guessing the code trying all the possible combinations (assuming there is no other way!).

How much time will it take you to guess the correct code?

Once again – How many different unlock codes could be possible?

Hmm.. That’s a tougher one!

Problems such as these require a systematic and a logical way of counting to be answered correctly, and, in a reasonable amount of time. This topic is all about that – to count without counting!

We’ll begin with the basic principles of counting, and as we proceed, we’ll develop some advanced techniques which can help us answer a lot of complex counting related problems and ultimately – count the uncountable!

Permutations & Combinations

Fundamental Principle of Counting

There are two basic principles of counting – The Multiplication Principle and The Addition Principle.

All subsequent concepts, (and formulas) in Permutations & Combinations will build upon these two principles, which are pretty simple to grasp.

To introduce the principles, let’s consider a simple problem:

Suppose a car company sells the following categories of cars…



Hatchback Sedan

Each of which comes in 3 different models… Standard, Deluxe and Sports

A simple question: How many different types of cars does the company sell?

Before reading further, think about it for a minute and try to find out the answer yourself (and of course, don’t worry about getting it wrong)

Let us list all the possible cases.

By counting all the cases, we can see that there are 6 different types of cars in all. (Did you get it right?)

Here is how we can reach that number by a simple calculation.

Since there are 2 car categories (hatchback & sedan), and each of these comes in 3 different models, the total number of types of cars can be obtained multiplying the number of categories (2) with the number of models available for each category (3), which is 2 x 3 = 6.

(In other words there will be 3 hatchback models + 3 sedan models = 2 x 3 = 6 total types of cars)

Let’s complicate the problem a bit. Suppose each of the above models were available in 2 different colours.

How many different cars do you think are available now? Let’s count again.

This time, there are 12 different types of cars in total.

Since each of the car is available in two colours, and there are 6 different types of cars (as calculated above), we have 6 blue cars + 6 black cars = 6 x 2 = 12 cars.

And this number can be obtained by multiplying together, the number of categories of cars (2), number of models for each category (3) and the number of colours available for each model (2): 2 x 3 x 2 = 12.

And what if there was another colour (say red) available for each model? The total number of cars would now be obtained by multiplying 2 (categories), 3 (models per category) and 3 (colour per model), which equals 18 (2 x 3 x 3).

(Note that this is 6 more than the number 12 obtained previously, as there would be 6 more red cars)

Make sense? Try listing these 18 cases to make sure everything is all right.



I’ll stop for now. In the next part, we’ll move on to the Multiplication Principle, and see some of its applications.

(PS: We’ve already used the Multiplication Principle thrice in this part!)

Multiplication Principal:

Now that you’ve got some idea of counting related problems and use of multiplication to answers them, I’ll now state the Multiplication Principle a bit formally:

Definition: “If a collection of objects can be separated into m different types, and each of these types can be separated into n different subtypes, then there will be m x n different types of objects in all.”

(There is a reason why I’ve been bolding each each along the way, because each each matters and things will be different without them. I’ll come to this a little later.)

To be more precise and define above multiplication principle mathematically,

Definition: If one experiment has n possible outcomes and another experiment has m possible outcomes, then there are m ´ n possible outcomes when both of these experiments are performed.

In other words, if a job has n parts and the job will be completed only when each part is completed and the first part can be completed in a1 ways, the second part can be completed in a2 ways and so on . . . the nth part can be completed in an ways, then the total number of ways of doing the job is a1a2a3...an. This is known as the Multiplication principle.

For our cars example, we could separate the cars into 2 types (sedan and hatchback) and each of these could be separated into 3 subtypes (models), making 2 x 3 = 6 total types of cars.

(We could also have gone the other way round and first separate the cars as models, and then further separate these into sedans and hatchbacks. The answer would have remained the same: 3 x 2 = 6)

Further, each of these 6 cars can be separated into 2 types (red and blue) making a total of 6 x 2 = 12 different types of cars.

The multiplication rule is not just limited to classification of objects. It can also be applied in different contexts. Consider this problem, for example – If there are 3 different flights from A to B and 2 different trains from B to C, in how many different ways can a person reach from A to C (using only these flights and trains)?

Here are all the cases for you to look at.

The possible routes are – 1a, 1b, 2a, 2b, 3a, and 3b – a total of 6.

Again, for each route from A to B, there are 2 routes from B to C. And since there are 3 routes from A to B, there will be a total of 3 x 2 different routes in all.

In other words, the number of ways will be obtained by multiplying the number of routes from A to B (3) with the number of routes from B to C (2) available for each route from A to B: 3 x 2 = 6.

Here’s another way we can state the multiplication principle:



“If a task T can be divided into subtasks T1 and T2, which can completed in m ways and n ways respectively, and T will be completed by completing both T1 and T2, then the number of ways of completing T will be m x n”

Let’s think of this example again.

The two subtasks T1 and T2 can be thought of as reaching from A to B and reaching from B to C respectively, where T1 can be completed in 3 ways and T2 can be completed in 2 ways.

Since the task of reaching from A to C will be completed by completing both T1 and T2, therefore the number of ways of reaching from A to C will be 3 x 2 = 6.

Here’s another simple problem.

A B C D

In how many ways can you reach from A to D, given the possible routes from A to B, B to C and C to D?

The task here is A to D, which will be completed by completing the tasks A to B and B to D.

Here B to D can further be divided into subtasks B to C (2 ways) and C to D (3 ways), both of which must be completed. Therefore the number of ways of reaching from B to D is 2 x 3 = 6.

And the number of ways of completing A to D will be 3 (A to B) x 6 (B to D) = 18 ways.

We could also directly write this number as 3 (A to B) x 2 (B to C) x 3 (C to D) = 18 ways.

To extend the Multiplication Principle:

““If a task T can be divided into n subtasks T1,T2,T3,….Tn, which can completed in m1, m2, m3… mn ways respectively and T will be completed by completing each of these subtasks, then the number of ways of completing T will be m1 x m2 x m3 x ……. x mn”

To elaborate above principle here are few solved examples, try to solve them yourself before seeing the solution.

Example1: A college offers 7 courses in the morning and 5 in the evening. Find the possible number of choices with the student if he wants to study one course in the morning and one in the evening.

Solution: The student has seven choices from the morning courses out of which he can select one course in 7 ways.

For the evening course, he has 5 choices out of which he can select one in 5 ways.

Hence the total number of ways in which he can make the choice of one course in the morning and one in the evening = 7 ´ 5 = 35.

Example 2: A person wants to go from station A to station C via station B. There are three routes from A to B and four routes from B to C. In how many ways can he travel from A to C?

Solution:



A ® B in 3 ways

B ® C in 4 ways

Þ A ® C in 3 ´ 4 = 12 ways

Remark:

The rule of product is applicable only when the number of ways of doing each part is independent of each other i.e. corresponding to any method of doing the first part, the other part can be done by any method.

Example 3: Find the number of three-digit natural numbers having digits in increasing order from left to right.

Solution: Here we have to fill three places.

First place can be filled by numbers 1, 2, ….7, second by 2, 3, ….8 and third by 3, 4, ….9 . i.e. the number of ways of filling each place is seven but total number of ways is not 7´ 7´7 = 343.

Reason behind this is that corresponding to 1 at first place, second place can be filled up by anyone of the seven digits 2–8 but when we put 2 at first place, the number of ways of filling second place is only six. So the number of ways of doing each part is not independent. So rule of product is not applicable in this case. The right approach for this problem is that first select three distinct non-zero digits which can be done in 9C3 ways, then arrange them in increasing order which can be done in one way only. Therefore, the required number of natural numbers is 9C3 ´ 1 = 84.

Example 4: How many (i) 5-digit (ii) 3-digit numbers can be formed by using 1, 2, 3, 4, 5 without repetition of digits.

Solution: (i) Making a 5-digit number is equivalent to filling 5 places.

Places:

Number of Choices:

5 4 3 2 1

The first place can be filled in 5 ways using anyone of the given digits.The second place can be filled in 4 ways using any of the remaining 4 digits.

Similarly, we can fill the 3rd, 4th and 5th place.

No. of ways of filling all the five places

= 5 ´ 4 ´ 3 ´ 2 ´ 1 = 120



Þ 120 5-digit numbers can be formed.

(ii) Making a 3-digit number is equivalent to filling 3 places.

Places:

Number of Choices:

5 4 3

Number of ways of filling all the three places = 5 ´ 4 ´ 3 = 60

Hence the total possible 3-digit numbers = 60.

Example 5: How many 4-letter words can be formed using a, b, c, d and e

(i) Without repetition (ii) with repetition.

Solution: (i) The number of words that can be formed is equal to the number of ways of filling the four places.

Places:

Number of Choices:

5 4 3 2

Þ 5 ´ 4 ´ 3 ´ 2 = 120 words can be formed when repetition is not allowed.

(ii) The number of words that can be formed is equal to the number of ways of filling the four places.

Places:

Number of Choices:

5 5 5 5

First place can be filled in 5 ways. If repetition is allowed, all the remaining places can be filled in 5 ways each.

Þ 5 ´ 5 ´ 5 ´ 5 = 625 words can be formed when repetition is allowed.

Fundamental Principle of Counting: Addition Principle

Hope that you now have some idea of the multiplication principle. This lesson will be focused on another basic principle of counting, known as the Addition Principle.



Let’s start with a simple problem: Suppose there are 3 different flights and two different trains connecting two places A and B. In how many ways can you reach from A to B (using only these flights or trains)?

The answer seems obvious: 5

Why? Because you can either choose one of the flights (3 choices) or choose one of the trains (2 choices). Therefore the total number of choices are 3 + 2 = 5 (Flight 1 or Flight 2 or Flight 3 or Train 1 or Train 2)

Another one: A restaurant offers 4 non-veg dishes and 3 veg dishes as starters. How many different types of starters does the restaurant offer?

Again, the answer will be 4 + 3 = 7.

So what is the addition principle about? Here is one way to state it:

“If two events A1 and A2 can occur in m and n ways respectively (none of these being common), then either of these events can occur in m + n ways.”

An analogous statement for the multiplication principle will be:

“If two events A1 and A2 can occur in m and n ways respectively, then both of these events can occur together in m x n ways.”

Now that we’ve some idea about the multiplication and the addition principles of counting, let’s dig a little deeper to understand them better.

Let us come back to the cars example again.

Suppose the Sports model exists only for the sedan and not for the hatchback. How many different types of cars are there now?

We can see that there are now 5 different types of cars, instead of 6 – 2 different hatchbacks and 3 different sedans.

In this case we cannot apply the multiplication principle directly, because each of the categories is not available in all the three models. Therefore we have to separate out the cases and count them individually.

Let’s bring in the colours now. Suppose all models are available in both blue and black. How many different cars would be there?

As you can see, there are 10 different cars.

In this case, we’ll have to use both the addition and the multiplication principles together.

There would be 2 x 2 = 4 hatchbacks, and 3 x 2 = 6 sedans, making a total of 10 different types of cars.

Or, total number of cars = (2 x 2) + (3 x 2) = 10.

Let’s complicate the problem a little more.



Suppose that the Deluxe and the Sports versions of the sedan category are also available in white. How many different cars are available now?

This time the total can be calculated as (2 x 2) + (1 x 2) + (2 x 3) = 12.

2 x 2 = 4 hatchbacks. And (1 x 2) + (2 x 3) = 8 sedans. A total of 12.

Thus, whenever the things to be counted need to be separated into cases, we count the cases separately and add them together.

To be more precise and define above multiplication principle mathematically,

Definition: If one experiment has n possible outcomes and another has m possible

outcomes, then there are (m + n) possible outcomes when exactly one of these experiments

is performed.

In other words, if a job can be done by n different methods and for the first method there are

a1 ways, for the second method there are a2 ways and so on . . . for the nth method, an ways,

then the number of ways to get the job done is (a1 + a2 + ... + an).

Fundamental Principle of Counting: Few Examples

This lesson will cover a few examples to help you understand better the fundamental principles of counting.

Question: Find the number of 3-digit numbers formed using the digits 3, 4, 8 and, 9, such that no digit is repeated.

Solution: The ‘task’ of forming a 3-digit number can be divided into three subtasks – filling the hundreds place, filling the tens place and filling the units place – each of which must be performed to complete the task.

Now there are 4 ways to complete the first subtask (i.e. filling the thousands place), since we have 4 digits to begin with. Next, there are 3 ways to complete to fill the tens place (the second subtask), as now there are only 3 digits left to be used (repetition of digits isn’t allowed). Lastly, there are 2 ways to complete the third task.

Therefore the total number of ways to complete the task (which equals the required number of 3-digit numbers) will be 4 x 3 x 2 = 24.

I’ll list out the 24 numbers, in case you don’t believe me.

348 349 384 389 394 398 438 439 483 489 493 498 834 839 843 849 893 894 934 938 943 948 983 984

Note that the digits could have been filled in any order, not necessarily in the order which I did. The answer obtained would have been the same.

Q2. How many 3-digit numbers can be formed using the digits 4, 5 and 6, such that the digit can repeat?

Soln. This seems similar. But this time the number of ways to fill the hundreds, tens and the units place would be 3 in each case, (and not 3, 2, 1) because the digits are allowed to be repeated.

Therefore the number of such 3-digit numbers would be 3 x 3 x 3 = 27. I’ll list them out again.



444 445 446 454 455 456 464 465 466 544 545 546 554 555 556 564 565 566 644 645 646 654 655 656 664 665 666

Okay. Let’s move on to a little more complex problems. I’ll stick to counting numbers in this lesson.

Q3. Find the number of 5-digit even numbers formed using the digits 1 to 9, such that repetition of digits is(i) allowed (ii) not allowed

Soln. I suggest that you grab a pen / paper and try to compute the answer yourself before reading further.

(i) In this case we have a restriction on filling the units place, as the number needs to be even. There are only 4 ways to do so (using 2, 4, 6 or 8). For the other four places, we have 9 choices each (as repetition is allowed).

Therefore the number of ways will be 9 x 9 x 9 x 9 x 4 = 32805

(I won’t list the numbers this time. You have no option but to believe me.)

(ii) This one is a little tricky. You might say that the answer is 9 x 8 x 7 x 6 x 4, as there will be 4 ways to fill the last digit (using 2, 4, 6 or 8).

But there’s a problem. We’re not sure that after filling the first four places (from the left), whether we’ll be left with 4 choices for the units place or not. Because one of the even digits might have already been used before we reach the units place.

To get around this problem, we fill the units place first using the even digits, followed by 8, 7, 6 and 5 choices respectively for the remaining places.

Therefore the number of such 5-digits numbers will be 4 x 8 x 7 x 6 x 5 = 6720.

Q4. Find the number of 4-digit odd numbers formed using the digits 0 to 9 such that repetition of digits is

(i) allowed (ii) not allowed

Soln. The units digit in this case can be filled in 5 ways (using the odd digits). Now, there are only 9 ways to fill the thousands place, as 0 cannot be used there. The remaining places can be filled 10 ways each.

Therefore the total numbers in this case would be 9 x 10 x 10 x 5 = 4500.

(Note that this is half of the total 9000 4-digit numbers possible, the other half being even)

(ii) We’ll first fill the units place with an odd digit – 5 ways. Next we have the thousands digit – 8 ways (leaving out the 0). Finally, 8 ways for the hundreds place (the 0 is back) and 7 ways for the tens place. The total number will be 5 x 8 x 8 x 7 = 2240

I guess that’ll be it for this lesson. As a practice problem try and count the number of 5-digit even numbers using the digits 0 to 9, with and without repetition.

Illustration 6: A college offers 7 courses in the morning and 5 in the evening. Find the number of ways a student can select exactly one course, either in the morning or in the evening.

Solution: The student has seven choices from the morning courses out of which he can select one course in 7 ways.



For the evening course, he has 5 choices out of which he can select one course in 5

ways.

Hence he has total number of 7 + 5 = 12 choices.

Example: A person wants to leave station B. There are three routes from station B to A and four routes from B to C. In how many ways can he leave the station B.

Solution:

B ® A in 3 ways

B ® C in 4 ways

Þ He can leave station B in 3 + 4 = 7 ways.

Another beautiful principle

Bijection Principle:

Generally it is not easy to count the numbers belonging to a very large collection (set) directly. Let us assume that we have a set X and we want to count n(x) ( number of elements in X). And we have another set Y and n(Y) can be counted (easily). Now if there exists one-one and onto relationship from set X to set Y, then n(X) = n(Y). In other words, if X and Y are two finite sets such that there is a bijection ( i.e. one-one onto map) from X to Y , then n(X) = n(Y). For example let us take a simple case. The sitting capacity of a class is 30. On a particular day, the class is full then, without actual counting, we can say that the total number of the students present in the class is 30.



Inclusion-Exclusion Principle:

Let U be a finite set of m elements and p1 , p2 , p3 , . . . ,pr be some properties which the objects of U may or may not have, then the number of elements of U which have at least one of the properties p1, p2, p3 , . . . ,pr is given by

= S1 - S2 + S3 – S4 + . . . +(–1)r-1Sr .and the number of elements of U which have none of the properties p1, p2, p3 , . . . ,pr is given by

= m – S1 + S2 - S3 + . . . + (-1) r Sr . Where Ai denotes the subset of U each of whose elements has the property pi( i = 1, 2, . . . ,

r) and denotes the subset of U none of those elements has the property pi

Ai Ç A j denotes the set of all elements of U having both properties p i and pj and so on. And Sk denotes the total number of elements of U having any k of the given r properties. E.g. S1 = n(A1) + n(A2) +...+ n(Ar), S2 = n(A1ÇA2) + n(A1Ç A3) +...+ n(A1ÇAr) + n(A2ÇA3) +...+ n(A2ÇAr) +...+ n(Ar-1 Ç Ar). In particular case:

For r =2, n(A1ÈA2) = n(A1) +n(A2) - n(A1ÇA2)

For r = 3, n(A1ÈA2 ÈA3) =n(A1) +n(A2) +n(A3) – n(A1ÇA2) – n(A1ÇA3) – n(A2ÇA3) + n(A1 Ç A2 ÇA3). For r = 4, n(A1ÈA2ÈA3ÈA4) = n(A1) +n(A2) +n(A3) + n(A4) – n(A1ÇA2) – n(A1ÇA3) – n(A1ÇA4) -n(A2ÇA3) – n(A2ÇA4) – n(A3ÇA4)+ n(A1 Ç A2 ÇA3) + n(A1 Ç A2 ÇA4) + n(A1 Ç A3 ÇA4) + n(A2 Ç A3

ÇA4) – n(A1ÇA2 Ç A3 ÇA4).

Example: Find the number of integers between 1 and 1000, both inclusive (a) Which are divisible by either of 10, 15 and 25. (b) Which are divisible by neither 10 nor 15 nor 25.

Solution: Let U denote the set of numbers between 1 and 1000 so that n(U) = 1000. Let A, B and C denote the sets of integers divisible by 10, 15 and 25 respectively.

Then n(A) = =100,

n(B) = = 66,

and n(C) = = 40 .S1 = n(A) +n(B) +n(C) = 100 + 66 + 40 = 206 Considering two of the three sets A, B, C, we have A Ç B = set of integers divisible by 10 and 15 = set of integers divisible by 30 (L.C.M of 10 and 15)

Þ n( A Ç B) = .Similarly, B Ç C = set of integers divisible by 15 and 25



= set of integers divisible by 75 (L.C.M of 15 and 25)

Þ n( B Ç C) = .Similarly, A Ç C = set of integers divisible by 10 and 25 = set of integers divisible by 50 (L.C.M of 10 and 25)

Þ n( A Ç C) = .S2 = 33 + 13 + 20 = 66.Finally, A Ç B Ç C = set of integers divisible by 10, 15 and 25 = set of integers divisible by 150 (L.C.M of 10, 15 and 25)

Þ n( A Ç B Ç C) = .S3 = 6 .

a) We have to find the number of integers divisible by either of 10, 15, and 25 i.e. n(AÈ BÈC) n(AÈ BÈC) = S1 – S2 +S3 = 206 – 66 + 6 = 146 .

b) The number of integers divisible by neither 10 nor 15, nor 25 = n(U) – n(AÈ BÈC) = 1000 – 146 = 854.

Factorials:

The factorial of a natural number n is the product of all natural numbers from 1 to n, that is, the product 1 x 2 x 3 x … x n. This product is denoted as n! (i.e. the number followed by an exclamation mark).

For example 4! = 1 x 2 x 3 x 4 = 24, and 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040.

As a convention, zero factorial (0!) is defined to be 1. (I’ll come back to this later). And you needn’t worry about factorial of negative numbers or fractions.

Note that n! = n x (n – 1)! For example 5! can be written as 5 x (4 x 3 x 2 x 1) which equals 5 x 4!

Okay, moving on.



Permutations

The term permutation is used to indicate ordered arrangements of objects. For example, the permutations of the three letters Q, W and E (in a row) are QWE, QEW, WEQ, WQE, EWQ and EQW.

The number of such permutations will be 3 x 2 x 1 = 3! = 6 (We’ve already seen the method of calculation in a previous lesson)

Similarly if we had four objects to be arranged in a row, for example, forming 4-digit numbers (without repetition) using 4, 6, 7, and 9, the number of permutations will be 4 x 3 x 2 x 1 or 4!

This number permutations of objects, taken all at a time (without repetition) is denoted as nPn or P(n, n). As you can see, this number comes out to be n! (n-factorial)In case we want the number of 4-digit numbers (without repetition of digits) using the digits 1 to 9, this will be 9 x 8 x 7 x 6.

This number is same as the number of permutations of 9 objects, taken 4 at a time. This is denoted as 9P4 or P(9, 4), which equals 9 x 8 x 7 x 6. The number can be expressed using factorials. If we multiply and divide it by 5 x 4 x 3 x 2 x 1, we get P(9, 4) = (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)/(5 x 4 x 3 x 2 x 1) = 9!/5! or 9!/(9 – 4)!. Neat.To generalize, the number of permutations of n different objects in a row, taken r at a time is denoted as nPr or P(n, r) which equals n!/(n – r)!Well, that’s it. Things will get better when you go through examples, which I’ll cover next. See you soon.

Permutations: Examples

This lesson covers a few examples relating to permutations, particularly those involving nPr. There isn’t be anything new to be covered here. The problems can be solved using the methods we’ve already seen. This is just to get you familiarized with factorials and notations.Q1. Find the number of 4-letter sequences that can be formed using the English alphabets, such that alphabets are not repeated in any sequence.

Sol. This equals the number of permutations of 26 different objects, taken 4 at a time – 26P4. We could also have computed the answer using the multiplication principle (as we did previously). The answer will come out to be 26 x 25 x 24 x 23 (which is same as 26P4)Q2. How many 4-digit numbers can be formed using the digits 3, 4, 5 and 7, using each digit only once?

Sol. The number of numbers in this case would equal 4P4 or 4! – number of permutations of 4 different objects, taken all at a time.Let’s complicate things a bit.

Q3. Find the number of 3-digit numbers formed using the digits 1 to 9, without repetition, such the numbers either have all digits less than 5 or all digits greater than 4.

Sol. Here, 342 is a possible number, 769 is another, but 385 is not. We can divide the numbers to be counted into two cases.

For those which have all the digits less than 5, we have to count the number of permutations of 4 digits (1, 2, 3, and 4), taken three at a time: 4P3.For those which have all the digits greater than 4, we have to count the number of permutations of 5 digits (5, 6, 7, 8 and 9), taken three at a time: 5P3.The total number of numbers would be 4P3+5P3, which equals 24 + 20 = 44.That’s it for now. Next, I’ll talk about circular arrangements of objects.

We can do the arrangement by box method also as I am describing as follows,


n-1 n-2 n-3n


(a) Arranging n objects, taken r at a time is equivalent to filling r places from n things.

r-Places:

Number of Choices:

The number of ways of arranging = The number of ways of filling r places

= n(n – 1) (n – 2) ….. (n – r + 1)

= = = nPr

(b) The number of arrangements of n different objects taken all at a time = npn = n!

With Repetition:

(a) The number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice…. upto r times in any arrangement

= The number of ways of filling r places where each place can be filled by any

one of n objects .

r-Places:

Number of Choices:

n n n n n

The number of permutations =The number of ways of filling r places

= (n)r

(b) The number of arrangements that can be formed using n objects out of which p are identical (and of one kind), q are identical (and of another kind), r are identical (and of

another kind) and the rest are distinct is .

Example: (a) How many anagrams can be made by using the letters of the word HINDUSTAN.

(b) How many of these anagrams begin and end with a vowel.

(c) In how many of these anagrams, all the vowels come together.

(d) In how many of these anagrams, none of the vowels come together. (e) In how many of these anagrams, do the vowels and the consonants

occupy the same relative positions as in HINDUSTAN.

Solution: (a) The total number of anagrams


n -(r-1)


= Arrangements of nine letters taken all at a time = = 181440.

(b) We have 3 vowels and 6 consonants, in which 2 consonants are alike. The first place can be filled in 3 ways and the last in 2 ways. The rest of the

places can be filled in ways. Hence the total number of anagrams = 3 ´ 2

´ = 15120

(c) Assume the vowels (I, U, A) as a single letter. The letters (IUA) H, D, S, T,

N, N can be arranged in ways. Also IUA can be arranged among themselves in 3! = 6 ways.

Hence the total number of anagrams = ´ 6 = 15120

(d) Let us divide the task into two parts. In the first, we arrange the 6

consonants as shown below in ways.

´ C ´ C ´ C ´ C ´ C ´ C ´ (C stands for consonants and ´ stands for blank spaces in between them)

Now 3 vowels can be placed in 7 places (in between the consonants) in 7p3 =

= 210 ways.

Hence the total number of anagrams = ´ 210 = 75600.

(e) In this case, the vowels can be arranged among themselves in 3! =

6 ways.

Also, the consonants can be arranged among

themselves in ways.

Hence the total number of anagrams = ´ 6 = 2160.

Example: How many 3 digit numbers can be formed using the digits 0, 1,2,3,4,5 so that

(a) digits may not be repeated

(b) digits may be repeated

Solution: (a) Let the 3-digit number be XYZPosition (X) can be filled by 1,2,3,4,5 but not 0. So it can be filled in 5 ways. Position (Y) can be filled in 5 ways again. (Since 0 can be placed in this

postion).



Position (Z) can be filled in 4 ways.

Hence, by the fundamental principle of counting, total number of ways is

5 x 5 x 4 = 100 ways.

(b) Let the 3 digit number be XYZ

Position (X) can be filled in 5 ways

Position (Y) can be filled in 6ways.

Position (Z) can be filled in 6 ways.

Hence by the fundamental principle of counting, total number of ways is

5 x 6 x 6 = 180.

Example: Find the number of ways in which 6 letters can be posted in 10 letterboxes.

Solution: For every letter, we have 10 choices (i.e. 10 letterboxes).

Hence the total number of ways = 106 = 1,000,000

Circular Permutations:

So far, we’ve only considered arrangements in a line or a row. This lesson will talk about arrangement of (distinct) objects in a circle.

Consider the problem: In how many ways can 5 persons be seated at a round table? Or, in how many ways can 5 distinct objects be arranged in a circle?

The arrangements differ only in the relative order of the objects, and the relative spacing between the objects does not matter.

To answer the question, consider a fictitious tire company called ANODE, which has to print its company name on its tires.

The five linear arrangements – ANODE, NODEA, ODEAN, DEANO and EANOD will be identical when arranged in a circular fashion. Have a look at the figure below.



Each of the tires can be rotated to align with any one of the rest. That means all of these arrangements are identical, even though we started with different linear arrangements.

Therefore 5 linear arrangements will correspond to only one circular arrangement.

Coming back to the question – In how many ways can 5 distinct objects be arranged in a circle?

Now there are 5! or 120 different linear arrangements possible. Therefore the number of circular arrangements will be 5! ÷ 5. (As every 5 linear arrangements will correspond to 1 circular arrangement). This is same as 4! or 24.

To generalize, the number of arrangements of n distinct objects in a circle will be n! / n, or (n – 1)!

Circular Permutations: Examples 4

Q1. In how many ways can 6 people be seated at a round table?

Soln. As discussed, the number of ways will be (6 – 1)!, or 120.

To the next..

Q2. Find the number of ways in which 5 people A,B,C,D,E can be seated at a round table, such that

(i) A and B must always sit together.(ii) C and D must not sit together.

Soln. (i) If we wish to seat A and B together in all arrangements, we can consider these two as one unit, along with 3 others. So effectively we’ve to arrange 4 people in a circle, the number of ways being (4 – 1)! or 6. Let me show you the arrangements:

But in each of these arrangements, A and B can themselves interchange places in 2 ways. Here’s what I’m talking about:

Therefore, the total number of ways will be 6 x 2 = 12.

(ii) The number of ways in this case would be obtained by removing all those cases (from the total possible) in which C & D are together. The total number of ways will be (5 – 1)! or



24. Similar to (i) above, the number of cases in which C & D are seated together, will be 12. Therefore the required number of ways will be 24 – 12 = 12.

Another example related to seating…

Q3. In how many ways can 3 men and 3 ladies be seated at around table such that no two men are seated together?

Soln. Since we don’t want the men to be seated together, the only way to do this is to make the men and women sit alternately. We’ll first seat the 3 women, on alternate seats, which can be done in (3 – 1)! or 2 ways, as shown below. (We’re ignoring the other 3 seats for now)

Note that the following 6 arrangements are equivalent:

That is, if each of the women is shifted by a seat in any direction, the seating arrangement remains exactly the same. That is why we have only 2 arrangements, as shown in the previous figure.

Now that we’ve done this, the 3 men can be seated in the remaining seats in 3! or 6 ways. Note that we haven’t used the formula for circular arrangements now. This is so because, after the women are seated, shifting the each of the men by 2 seats, will give a different arrangement. After fixing the position of the women (same as ‘numbering’ the seats), the arrangement on the remaining seats is equivalent to a linear arrangement.

Therefore the total number of ways in this case will be 2! X 3! = 12.

I hope that you now have some idea about circular arrangements. Next, I’ll talk about Combinations.

The previous lessons talked about arrangements or permutations of objects. This lesson will deal with selections or combinations of objects. In simple words, a combination is a selection made with no regard to order of the selected objects.

To be a little more clear, let’s take an example. Suppose you have 4 different colors…


http://doubleroot.in/wp-content/uploads/2014/01/pc7-circular-permutations-examples-31.png


…of which you have to mix 2 (in equal proportions) to make a new color. How many new colors can be made?

Notice that mixing Blue and Red is exactly the same as mixing Red and Blue, i.e. the order does not matter.

Let’s list all the possible cases.

As you can see there are 6 new colors – Black-Blue, Black-Red, Black-Yellow, Blue-Red, Blue-Yellow, Red-Yellow.

Okay, so how do we count this in a logical way?

Let’s first state the problem in a more standard manner: “In how many ways can a selection of 2 objects be made from 4 distinct objects?” (say A, B, C and D)

Let the number of combinations be N.

Note that every selection of two particular objects (say A and B) corresponds to 2! different arrangements (AB and BA). That is, there are 2! times as many permutations as there are combinations (for the case of selecting 2 objects).



But we’ve already calculated the number of permutations of 4 objects taken 2 at a time. This equals 4P2. Therefore, according to our logic, 4P2 = 2! X N, or N = 4P2/2! = 6. Makes sense?What if we had to select 3 objects out of 5? In this case the number of permutations will be 3! or 6 times as the number of combinations. (The 6 permutations ABC ACB BAC BCA CAB CBA correspond to the selection {A, B, C}). Therefore the number of combinations will be the number of permutations divided by 3!, which equals 5P3/3! = 5!/(2!3!) = 10. (These 10 selections are {A,B,C}, {A,B,D},{A,B,E},{A,C,D},{A,C,E},{A,D,E},{B,C,D},{B,C,E},{B,D,E},{C,D,E})We can now generalize this expression for the case of selection of r objects out of n distinct objects (0 ≤ r ≤ n). The number of selections or combinations will be nPr / r! or n!(n−r)!r!. This expression is denoted as nCr or C(n, r) or (nr). (We’ll stick with the first one)Note that each object can be selected exactly once. Later we’ll come to the case when we have to select r objects out of a lot n distinct types of objects (each available in unlimited supply), such that an object of a given type can be selected more than once.

To summarize: The number of ways of selecting r objects out of n distinct objects is given by nCr, which equals n!(n−r)!r!Before moving on to examples, I’ll show you one more way to arrive at the same expression for selections.

Consider 6 different objects: A, B, C, D, E and F

Now suppose we have to select 1 object from the lot. Clearly, we have 5 different possible selections: A, B, C, D, E or F. This is same as 6C1 as we derived in the previous lesson.Okay, what if we have to select 2 objects? First we’ll select any 1 object in 6C1 ways. Now there are 5 objects remaining, of which we have to select 1 more. This can be done in 5 ways. So the total number of ways to complete the selection will be 6C1 x 5.But there is a problem. In the above way of calculating, we’ve counted “selecting A followed by B” and “selecting B followed by A” as different. But they are the same, as the order of selection does not matter. We’ve counted the same selection twice.

Therefore to arrive at the correct answer we’ll have to divide the previous answer by 2. Therefore the total number of selections will be (6C1 x 5)/2.This is same as 6!5!1!×52=6!4!2!=6C2.What about 3 objects? First, select any 2: 6C2 ways. Then out of the remaining 4, select any 1: 4 ways. Now again if we multiply 6C2 by 4, we would have counted each selection thrice (A,B followed by C; A,C followed by B; B,C followed by A). So we’ll have to divide the product by 3: (6C2×4)/3=6!2!4!×43=6!3!3!=6C3. Same answer as we derived earlier!I think now we can generalize this logic to n objects, inductively. If we know the way to select ‘r-1’ objects the number of way to select ‘r’ objects will be obtained by multiplying the previous by (n – (r – 1)) (as there will be these many objects left after we’ve selected ‘r – 1’, and we’ve to select 1 more) and then dividing by r (as each selection will be repeated r times, as we saw above)

The number of ways to select 1 object = n or nC1.The number of ways to select 2 objects will be nC1×n–12=n!1!(n–1)!×n–12=n!2!(n–2)!=nC2The number of ways to select 3 objects will be nC2×n–23=n!2!(n–2)!×n–23=n!3!(n–3)!=nC3The number of ways to select 4 objects will be nC3×n–34=n!3!(n–3)!×n–34=n!4!(n–4)!=nC4And so on… I hope you get the idea.

We’ve reached the same expression/formula – nCr – to obtain the number of selections of r objects from n distinct objects, using a different logic. (There might be many more.)

Now you’ll learn a result about binomial coefficients: nCr=nCn−r, using the formula as well as by using combinations.



For example 7C2 equals 7C7−2 or 7C5.Lets calculate to be sure.

7C2=7!2!(7−2)!=7.6.5.4.3.2.1(2.1)(5.4.3.2.1)=21 and 7C5=7!5!(7−5)!=7.6.5.4.3.2.1(5.4.3.2.1)(2.1)=21Observe that the “2!” and the “5!” just got exchanged, while the numerator remains the same. Can you see why this will hold true for all kinds of nCrs in general ?If not, here’s why. Using the formula for nCr, which equals n!r!(n−r)!, we have nCn−r=n!(n−r)!(n−(n−r))!=n!(n−r)!(r)!=n!r!(n−r)!=nCrIn other words, the r! and (n-r)! get interchanged, and the expression’s value remains the same.

Okay, now there’s another way we can reach the same result. Recall that nCr is the number of ways to select r objects from n different objects.To prove that nCr=nCn−r, we have to argue that the number of ways to select r objects from n different objects should be the same as the number of ways to select n – r objects from n different objects.As an example, suppose we are to select 3 letters from out of the 4 letters A,B,C and D. To select the letters, we line up the 4 letters, and make the selections by circling out any 3 of these 4 letters. Here are the possible selections..

There are 4 possible selections (which is same as 4C3, as derived earlier). Now to make the same selections, we could do the following instead..



That is, put a cross mark on that letter, which is not to be selected. And in how many ways can we do that? In 4C1 ways! Why? Because out of 4 letters you have to select 1 (and put a cross on it).So the selection of 3 letters out of 4 can be done by putting a circle each on 3 of the letters to be selected OR by putting a cross on the letter not to be selected. That is, selecting 3 objects (out of 4) is exactly the same as selecting 1 object (and throwing it away, thereby selecting the remaining 3 objects).

Well? This means that 4C3 must equal 4C1, as we’re essentially doing the same thing in both these methods (which should make the number of ways of both the methods equal)To generalize, selecting r things out of n different things is exactly same as selecting n – r things and removing them from the lot, so the remaining r things get automatically selected. Therefore, the number of ways to the former, i.e. nCr must equal the number of ways to do the latter nCn−rAnd phew ! We’re done ! nCr=nCn−rTo summarise: Suppose n persons (a1, a2, a3,…,an) are to be arranged around a circular table. There are n! ways in which they can be arranged in a row. On the other hand, all the linear arrangements depicted by

a1, a2, a3, ……..., an

an, a1, a2,……….,an – 1

an – 1, an, a1, a2…..an – 2

…………………….

…………………….

…………………….

a2, a3, a4,……….,a1

will lead to the same arrangement for a circular table. Hence each circular arrangement corresponds to n linear arrangements (i.e. in a row). Hence the total number of circular

arrangements of n persons is = (n – 1)!.



In other words, the arrangement (permutation) in a row has a beginning and an end, but there is nothing like beginning or end in circular permutation. Thus, in circular permutation, we consider one object as fixed and the remaining objects can be arranged in (n – 1)! ways (as in the case of arrangement in a row).

Distinction between clockwise and anti-clockwise Arrangements:

Consider the following circular arrangements:

In figure I, the order is clockwise whereas in figure II, the order is anti-clock wise. These are two different arrangements. When distinction is made between the clockwise and the anti-clockwise arrangements of n different objects around a circle, then the number of arrangements = (n – 1)!

But if no distinction is made between the clockwise and the anti-clockwise arrangements of n

different objects around a circle, then the number of arrangements is (n – 1)!

For an example, consider the arrangements of beads (all different) on a necklace as shown in figures A and B.

Look at (A) having 3 beads x1, x2, x3 as shown. Flip (A) over on its right. We get (B) at once. However, (A) and (B) are really the outcomes of one arrangement but are counted as two different arrangements in our calculation. To nullify this redundancy, the actual number of different arrangements is (n-1)!/2.

Remarks:

When the positions are numbered, circular arrangement is treated as a linear arrangement.

In a linear arrangement, it does not make difference whether the positions are numbered or not.

Example: Consider 23 different coloured beads in a necklace. In how many ways can the beads be placed in the necklace so that 3 specific beads always remain together?

Solution: By theory, let us consider 3 beads as one. Hence we have, in effect, 21 beads, 'n' = 21. The number of arrangements = (n-1)! = 20!

Also, the number of ways in which 3 beads can be arranged between themselves is 3! = 3 x 2 x 1 = 6.

Thus the total number of arrangements = (1/2). 20!. 3!.

Example: In how many ways 10 boys and 5 girls can sit around a circular table so that no two girls sit together.



Solution: 10 boys can be seated in a circle in 9! ways. There are 10 spaces inbetween the boys, which can be occupied by 5 girls in 10p5 ways. Hence total number of ways

= 9! 10p5 = .

Example: A question paper consists of 10 questions of which a student needs to answer any 7. In how many ways can the student make his selection?

Solution: This is a simple case of selection of 7 objects (questions) out of 10 distinct objects. The number of ways will be 10C7 = 120

Example: Find the number of ways in which a team of 4 can be selected from a group of 10 people A1, A2, A3, …, A10 such that

(i) there are no restrictions(ii) A1 and A3 must be selected(iii) A2 must not be selected(iv) if A6 is selected then A4 must also be selected(v) if A5 is selected then A7 must not be selected(vi) A9 is selected if and only if a A10 is selected

Solution: Let’s take ‘em one by one!

(i) This one’s plain: 10C4

(ii) In this case, we must select A1 and A3. Done. Now, out of the remaining 8, we’ve to select 2 more. The number of selections will be 8C2

(iii) In this case, we’ll remove A2 from the group, and select any 4 from the remaining group of 9 people – in 9C4 ways.

(iv) This one is a little complicated. We’ll have to divide our counting into 2 cases. First in which A6 is selected, and the second in which A6 is not selected.

Case 1: According to the restriction, if A6 is selected then A4 also must be selected. So we’ll select these 2, plus 2 more from the remaining 8, in 8C2ways.

Case 2: If A6 is not selected, then we can select any 4 out of the remaining 9, in 9C4 ways.

Since it will be either Case 1 or Case 2, the number of ways in both these cases will be added (the addition principle). The total possible selections will be 8C2 + 9C4.



(v) Again, we’ll divide our counting into 2 cases. First in which A5 is selected, and the second in which A5 is not selected.

Case 1: According to the restriction, if A5 is selected then A7 must not be selected. So we’ll remove A7, and select 3 more from the remaining 8: 8C3ways.

Case 2: If A5 is not selected, then we can select any 4 out of the remaining 9, in 9C4 ways.

The total possible selections will be 8C3 + 9C4.

(vi) In this case, either both A9 and A10 will be selected, or both will not be selected.

Case 1: Select A9 and A10, and select 2 more from the remaining 8: 8C2 ways

Case 2: Remove A9 and A10 from the group, and select 4 from the remaining 8: 8C4ways

Total number of ways: 8C2 + 8C4

Let’s now complicate things a bit more..



Example: In how many ways can a committee of 8 be selected from a group of 10 men and 12 women, such that, in the committee,

(i) there are 3 men and 5 women;(ii) there are atleast 6 women;(iii) there is atleast one man;(iv) there are atmost 2 women;(v) there are more men than women ?

Solution: In this problem, we’ll divide our task of selection into 2 subtasks – (a) select the men and (b) select the women. Since both of these need to be completed to complete the selection, we’ll multiply the number of ways obtained in both these cases (the multiplication principle)

(i) The men can be selected in 10C3 ways, and the women can be selected in 12C5 ways. Therefore the total number of ways will be 10C3 x 12C5.

(ii) Now in this case, there can either be 6 women (and 2 men), or 7 women or all 8 women.

We’ll therefore count the three cases separately and add them together. Similar to (i) above, when there are 6 women (and 2 men), the number of cases will be 10C2 X 12C6. And when there are 7 women (and 1 man) the number of cases will be 10C1 x 12C7. Finally in the case of 8 women and 0 men, the number of selections will be 10C0 x 12C8, or simply 12C8. The final answer becomes 10C2 x 12C6 + 10C1 x 12C7 + 12C8

(iii) This could be counted in a similar manner as (ii) above. But there is a better method. If from all the possible selections, we remove those in which no man is selected (or all-woman committees), we’ll be left with those cases in which there is atleast one man. Sounds right?

The total number of 8-member committees will be 22C8. The total number of all-women committees will be 12C8. Therefore our required answer will be 22C8 – 12C8. Pretty neat!

(iv) Either all men, or 7 men and 1 woman, or 6 men and 2 women: 10C8 + 10C7 x 12C1 + 10C6 x 12C2

Combinations: More Examples

. Let’s begin with few more examples...

Q1. There are 10 points in a plane, no 3 of which are collinear. How many different lines can be formed by joining these points?

Soln. To form a line, all we need to do is select any 2 points (subtask 1), and then join them (subtask 2). The number of ways to select any points (out of 10 distinct points) will be 10C2. Once we select the points, there is only 1 straight line which will be formed using these points. Therefore the number of lines will be 10C2 x 1 or 45. Here’s a crazy figure to illustrate.



Q2. Suppose 4 of the 10 points in Q1 are collinear, and no three of the remaining are collinear. Find the number of straight lines which can be formed. Also find the number of triangles which can be formed.

Soln. Since 4 particular points are now collinear, there will be only 1 line which will be formed by joining any 2 of these 4, instead of 4C2 if they had been non collinear. So, we’ve lost 4C2 or 6 lines, and instead we got only 1. Therefore, the total number of lines will be 10C2 – 4C2 + 1 = 40

There is one more way in which we can calculate the number of lines. Let’s divide the points in two groups: the collinear group of 4 points, and the non-collinear group of 6 points.

To count the number of lines, we have three possible cases. First, the lines formed using the points of the collinear group – only 1 line. Second, the lines formed using only the points of the non-collinear group – 6C2 or 15 lines. Third, select one point from the collinear group (4C1 ways) and the other from the non-collinear group (6C1 ways) – the number of such lines will be 4C1 x 6C1 = 24.

Therefore the total number of lines will be 1 + 15 + 24 = 40. This was a bit longer method, but nevertheless an important method of counting – dividing into subcases and adding the results together. Another crazy figure..



What about the number of triangles?

Suppose all of the points were non-collinear. The number of triangles would have been 10C3, because to form a triangle, we need to select any three points.

But! There are four culprit points, which won’t form a triangle, as they are collinear. We’ve lost 4C3 triangles, which would have been formed, had these 4 been non-collinear.

Therefore, the total number of triangles is 10C3 – 4C3.

Try using the second method (of making cases) and see if you arrive at the same answer.

Q3. Consider a card game, in which each player is dealt 5 cards. In how many ways can a player obtain the following combinations?

A deck of cards contains 52 cards with 4 different suits, and 13 kinds of cards in each suit, as shown below.

A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠

A♥ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥

A♦ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ J♦ Q♦ K♦

A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣

(i) All cards of the same suit:

3♥ 7♥ 8♥ 9♥ K♥

(ii) Two cards of the same kind and the rest three of the same kind:



4♥ 4♠ K♥ K♣ K♦

(iii) Two cards of the same kind and rest all different:

4♥ 4♠ K♥ 9♣ 7♦

(iv) Cards in sequence:

4♥ 5♠ 6♥ 7♣ 8♦

Soln. Note that the order of the cards in a player’s hand is irrelevant. We are only interested in the combination of cards.

(i) If we need 5 cards of the same suit, we’ll first have to select a suit, in 4C1 ways. We now have a suit of 13 different cards, of which 5 cards need to be selected. This can be done in 13C5 ways. Therefore the total number of ways will be 4C1 x 13C5 = 5148

(ii) We’ll divide this task into the following subtasks – select the kind to be paired, then select 2 cars of that kind. Then select the kind to be tripled, then select 3 cards of that kind.

For the pair we first select the kind in 13C1 ways, and then select any two cards of that kind in 4C2 ways.

We are now left with 48 cards, 12 kinds of cards of 4 different suits. Now we’ll select another kind in 12C1 ways and select 3 cards of that kind in 4C3ways.

The total number of combinations is therefore 13C1 x 4C2 x 12C1 x 4C3 = 3744

(iii) Similar to the previous part we can select the pair in 13C1 x 4C2 ways. But now we have to be careful. Of the 48 cards remaining, we cannot select any 3 cards, because it might happen they all are of the same kind.

To handle this, we’ll first select 3 different kinds, in 12C3 ways, say we get 5, 8 and K. Now each of these cards can be of any of the 4 suits. There are 4 possibilities for the ‘5’, 4 for the ‘8’ and 4 for the ‘K’. That means, for a given selection of 3 kinds of cards, there are 4 x 4 x 4 or 64 different combinations.

Therefore, the total number of combinations of the remaining 3 cards will be 12C3 x 64, making the answer equal to 13C1 x 4C2 x 12C3 x 64 = 1098240

(iv) In this case we’ll first select the sequence to be formed. There are 9 possibilities: (A,2,3,4,5), (2,3,4,5,6) … (9, 10, J, Q, K). Similar to the previous part, there will be 4 possibilities for each of the 5 cards selected. Therefore the total number of combinations will be 9 x 4 x 4 x 4 x 4 x 4 = 147456

Perplexed? Me too! Try a few more cases yourself..

(v) Two pairs, and the fifth card different

4♠ 4♥ 5♣ 5♦ K♦

(vi) Three of the same kind, and the rest two different

7♠ 7♥ 7♣ 4♦ K♣

(vii) Four of the same kind

4♠ 4♥ 4♣ 4♦ K♦



Permutations involving Identical Objects 33

So far we’ve talked about permutations of objects which were distinct. Things change when some (or all) of the objects to be arranged are identical.

Suppose we have to arrange 5 balls in a row of which 3 are of the same color, and the other two are of different colors. How many different arrangements are possible?

Here’s one way to think about this problem.

Let’s designate five boxes, each of which is to be filled by one ball.

First, we’ll place the identical balls in any three of the five boxes. This can be done in 5C3 ways. Here’s one of them.

Next, in the remaining two boxes, we’ll place the remaining two balls. This can be done in 2! ways.

or



Therefore the total number of arrangements will be 5C3 x 2! = 5!/3! = 20.

Now let’s say 3 of the 5 balls are of the same colour and the other 2 are of different colours. How many different arrangements are possible?

Similar to the method used above, we’ll first place the identical balls in any 2 of the 5 boxes, in 5C2 ways. One of the arrangement is given below.

Now in the remaining three boxes, the remaining three balls can be placed in 3! ways.

Therefore the total number of arrangements will be 5C2 x 3! = 5!/2! = 60.

We could also have placed the different coloured balls first. This can be done by first selecting any three boxes, (in 5C3 ways), and then placing the balls in the boxes (3! ways).

After we’ve placed these three balls, the remaining two balls will be filled with the two identical balls (which can be done in only one way).

The answer remains the same (5C3 x 3!) x 1 = 5!/2! = 60

Do you see a pattern? Had the 5 balls been all of different colours, the total number of arrangements would have been 5!. When 3 of the balls were identical, the answer changed



to 5!/3!, and when 2 of them were identical, the answer was 5!/2!. Can we arrive at a formula or a generalization?

Suppose there are n objects, to be arranged in a row, of which p objects are identical, and the remaining are distinct. How many arrangements are possible?

In this case we have n boxes, each to be filled with one ball. We’ll first select any p boxes out of the n boxes, in nCp ways. Now we’ll place the identical balls in the selected boxes, one in each. We’re now left with n – p boxes and n – p balls, which can be placed in the boxes in (n – p)! ways.

Therefore the total number of ways is nCp (n – p)! or n!/p!

Permutations & Combinations – Permutations involving Identical Objects: Examples

This lesson will cover a few examples to illustrate what we learnt in the previous lesson.

Q1. Find the number of arrangements (in a row) of the letters of the following words:

(i) TWITTER

(ii) GOOGLE

Sol. (i) There are a total of 7 letters to be arranged in a row of which 3 T’s are identical (and the remaining different). The total number of permutations will be 7!/3! or 840.

(ii) In this case, there are a total of 6 letters which 2 G’s are identical and 2 O’s are identical. The number of arrangements will be 6!/2!2! or 180

Q2. Consider the following grid

Here is the situation. You are at the bottom left corner of the grid (the red dot), and are supposed to reach the top right corner of the grid. But there is a condition – you are only allowed to travel along the grid lines, either one step rightward or one step upward. In how many different ways can you reach the top right corner? One such way is shown below.



Sol. This is one example of problem where it isn’t very obvious whether it’s based on permutations or combinations. You’ll come across many such problems, which won’t involve direct application of any formula. In such problems, you need to establish a logic for counting first, and accordingly use the basic counting principles (or formulas, if applicable).

Let’s come back to the problem again. Notice that whatever be the path chosen, there will always be a total of 5 steps in the rightward direction and 4 in the upward direction. Here are a few examples.



So how does one path differ from another? Only in the order of these forward and upward steps taken.

Let’s denote each path by a series of U’s (upwards) and R’s (rightwards). Each series will consist of 5 R’s and 4 U’s in some order, and will correspond to a unique path.

For example, the first path will be denoted by UUUURRRRR, the second by URUURRRUR, and the third by URRURURUR.

So what about the number of paths? To count that, we’ll instead count the number of these letters, as each arrangement corresponds to a unique path.

Now we have 9 letters, of which 4 U’s are identical, and 5 R’s are identical. The number of arrangements and therefore the number of paths will be 9!/4!5!

Neat!

To count the paths, we counted the number of arrangements of letter series, which was easier to think about and count. Many P & C problems will be about establishing such links between two countable things, which will make things much easier. I’ll cover some more examples of this kind later on.

As an exercise try the following variant of the above problem on your own.

Q. Consider the same grid as in the previous problem. Starting again at the bottom left, how many different paths to the green dot are possible, if you must pass through the blue dot every time?



Permutations & Combinations – Combinations involving Identical Objects

This lesson will cover briefly a few simple cases involving selections (or combinations) involving identical objects.

Suppose you have five identical red balls, of which you’ve to choose any two. How many different combinations are possible?

It doesn’t matter which two balls you select – whether you select the first two, or the last two, or the third and the fourth – you’ll end up with the same selection each time. Why? Because the balls are identical.

Therefore the number of ways to select two balls will be one.

And if you had to select three balls, how many choices would you have? The answer remains the same – one.

And four balls? Again, only one possible selection.

To formalize things a bit –

The number of ways to select r objects out of n identical objects equals 1 (where 0 ≤ r ≤ n)

Okay, that was simple to understand. Let’s complicate things a bit now.

Suppose you now have the following lot – three red balls and three balls of different colors

If you’re to select three out of these, how many different choices do you have?

The answer again not be 6C3 (the case where all of the balls are of different color), because identical balls will reduce the number of choices, as we saw above.



So how can we count the possible combinations in this case? By making cases

We first separate the balls into two lots – the identical balls (say, lot 1) and the distinct balls (lot 2).

Next, we divide our selection into two subtasks – select from lot 1 and select from lot 2.

Finally, we make cases..

I. All the three balls from lot 1: 1 way

II. Two from lot 1 and one from lot 2: 1 x 3C1 ways

III. One from lot 1 and two from lot 2: 1 x 3C2 ways

IV. All the balls from lot 2 – 3C3 ways

.. and we add all the cases: 1 + 3 + 3 + 1 = 8 ways

Hmm... It’s getting complicated already.

And you must be thinking, “Wait.. what about the case where we’ve to select 3 balls from a lot of 4 red, 5 green and 6 different coloured balls?”

Exponent of Prime p in n!:

Let p be a prime number and n be a positive integer. Then the last integer amongst

1, 2, 3, …….., (n –1), n which is divisible by p is p, where denotes the greatest

integer less than or equal to .

For Example, = 3, = 2, = 5 etc.

Let Ep (n) denotes the exponent of the prime p in the positive integer n. Then,

Ep (n !) = Ep (1 . 2. 3 …..(n –1) n)

= Ep [ Remaining integers between 1 and n are not divisible by p]

= + Ep

Now, the last integer amongst 1, 2, 3, ….. which is divisible by p is

\ Ep (n !) = + + Ep .



Continuing in this manner, we get

Ep (n !) = + + + ……..+ , where s is the largest positive integer such that ps £ n < ps + 1

Example: Find the exponent of 3 in 100 !

Solution: Let Ep (n) denotes the exponent of p in n.

Then, Ep (n !) = + + + ……..+ , where s is the largest positive integer such that ps £ n < ps + 1

Here, n = 100, p = 3 34 £ 100 < 35 \ s = 4.

So, E3 (100 !) = + + +

= 33 + 11 + 3 + 1 = 48

Hence, the exponent of 3 in 100 ! is 48.

Example: Find the number of zeros at the end of 100 !.

Solution: In terms of prime factors 100! can be written as 2a3b5c, 7d………..

Now,

E2 (100!) =

= 50 + 25 + 12 + 6 + 3 + 1 = 97

and E5 (100 !) = = 20 + 4 = 24.

Therefore,

100 ! = 297 ´ 3b ´ 524 ´ 7d ´ …….= 273 ´ (2 ´ 5)24 ´ 7d ´………

= 1024 ´ 273 ´ 3b ´ 7d ´……….

Thus, the number of zeros at the end of 100 ! is 24.



Remarks:

The number of integers from 1 to n which are divisible by k is . e.g. the number

of integers from 1 to 100 which are divisible by 2 and 3 individually are

and = 33 respectively.

The number of natural numbers from 1 to n, which are perfect kth powers, is

[n1/k]. e.g. the number of perfect squares from 1 to 110 is [ ] = 10.



ASSIGNMENTS (Subjective)

LEVEL – I

1. Find the total number of words which can be formed out of the letters of the word ALLAHABAD such that the vowels occupy the even positions.

2. On a railways there are 20 stations. Find the number of different tickets required in order that it may be possible to travel from every station to every station

3. A car will hold two persons in the front seat and one in the rear seat. If among six persons only two can drive. Find the number of ways in which the car can be filled.

4. The letters of the word CRICKET are written in all possible orders and then are arranged in a dictionary what is the rank of the word CRICKET.

5(a). A cricket team of eleven players is to be formed from 16 players including 4 bowlers and 2 wicket keeper. In how many different ways can a team be formed so as to contain atleast 3 bowlers and atleast one wicket keeper.

(b). In how many ways can a team of 3 chemistry teachers and 4 mathematics teachers can be formed from 8 chemistry teachers and 10 mathematics teachers such that a particular chemistry teacher refuses to be in the team if a particular mathematics is in the team.

6. A round table conference is to be held between 20 delegates of 20 countries. In how many ways can they be seated if two particular delegates

(a) always sit together.

(b) never sit together.

7. (a) In how many ways can 2n people be divided into n pairs ?

(b) In how many ways can you equally distribute 100 packages of food to 10 refugees?

(c) In how many ways mn things be equally distributed in n groups?(d) In how many ways can 22 books be divided into 5 students so that two students will have 5 books each and the other three students will have 4 books each.



8. There are 5 mangoes and 4 apples. In how many different ways can a section of fruits be made if

(a) fruits of the same kind are different.

(b) fruits of the same kind are identical?

9. Five subjects are to be allotted to 6 different periods. In how many ways can this be done if no period has to go un allotted?

10. In how many ways can 3 girls and 9 boys be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many sitting arrangements are possible if 3 girls should sit together in the back row on adjacent seats?



LEVEL – II

1. How many numbers of 4 digits can be made with the digits 0, 1, 2, 3, 4, 5, which are divisible by 3, digits being unrepeated in the same number? How many of these will be divisible by 6?

2. In how many ways can 21 identical white balls and 19 identical black balls be arranged in a row so that no two black balls may be together? What will be the result if balls are considered to be distinct?

3(a). A man has seven relatives, four women and three men. His wife also has seven relatives, three women four men. In how many ways they invite three women and three men so that three of them are the man’s relatives and three his wife’s.

(b). A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw.

(c). Find the total number of seven digit numbers the sum of whose digit is odd.

4. A person wants to hold as many different parties as he can out of 24 friends, each party consisting of the same number. How many should he invite at a time ? In how many of these would the same man be found ?

5. There are n straight line in a plane, no two of which are parallel and no three pass through the same point, their points of intersection are joined. Show that the number

of fresh lines thus introduced is

18 n(n –1)(n –2)(n –3).

6(a). Suppose n different games are to be given to n children. In how many ways can this be done so that exactly one child gets no game.

(b). Find the numbers of rectangles excluding square from a rectangle of size 15 ´ 10.



7. In how ways can 10 persons take seats in a row of 24 fixed seats so that no two person take consecutive seats?



8. In how many ways can the letters in the English alphabet be arranged, so that there are 7 letters between the letters a and b.

9(a). A mathematics paper has 12 questions divided into 3 sections. A, B, C, each having 4 questions. In how many ways can you answer 5 questions selecting at least one question from each part.

(b). In an examination, the maximum marks for each of 3 papers is n and for fourth paper 2n. Prove that the number of ways in which a candidate can get 3n marks is 16 (n + 1) (5n2 + 10n + 6).

10. From a given number of 4n books, there are three sets of n identical books on physics, chemistry and mathematics. The remaining n are distinct books on other subjects. Find the number of ways of choosing n out of 4n books.

ANSWERS (Subjective)

LEVEL – I

1. 60 2. 380

3. 40 4. 531

5(a). 2472 (b). 8C3 ´ 10C4 - 7C2 ´ 9C3

6(a). 2 ´ 18! (b). 17 ´ 18!

7(a).

2n !

n ! 2n (b).

100 !

(10 !)10

(c).

(mn) !(m! )n n ! (d).

22 !

3! 2! (5 !) ( 4 ! )3

8(a). 29 – 1 (b). 29

9. 1800 10. 14P12, 24 ´ 11P9



LEVEL– II

1. 96, 52 2. 1540, 1540 ´ 21! ´ 19!

3(a). 485 (b). 64

(c). 4500000

4. 12 friends, 23C11 parties

6(a). nC2 ´ n! (b). 5940

7.

15 !5 ! 8. 36 ´ 24!

9(a). 56 10. 2n –3(n2 + 7n + 8)



ASSIGNMENTS (Objective)

LEVEL - I

1. The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive, is

(A) 4P4 (B) 4P4 ´ 3P3

(C) 6P6 ´ 3P3 (D) 3P3 ´ 3P3

B

2. Every one of the 10 available lamps can be switched on to illuminate certain hall. The total number of ways in which the hall can be illuminated is (A) 55 (B) 1023

(C) 210 (D) 20!

B

3. Value of expression 47C4 + ∑j=1

552− jC3

is equal to

(A) 52C4 (B) 52C2

(C) 52C6 (D) none of these

A

4. A polygon has 65 diagonals. The number of its sides is

(A) 8 (B) 10

(C) 11 (D) 13

D

5. Set A has 3 elements and set B has 4 elements, the number of injections that can be defined from A to B, is

(A) 144 (B) 12

(C) 24 (D) 64

C

6. The letter of the word ‘MATHS’ are written in all possible orders and these words are written out as in dictionary, then the rank of the word ‘MATHS’ is

(A) 53 (B) 54

(C) 55 (D) 56



A

7. Number of ways of painting the faces of a cube with different colours

(A) 1 (B) 6

(C) 6! (D) 26

A

8. The maximum number of different permutations of 4 letters of the word ‘EARTHQUAKE’ is

(A) 1045 (B) 2190

(C) 4380 (D) 2348

B

9. Everybody in a room shakes hand with everybody else. The total numbers of handshakes is 66. The total number of persons in the room is

(A) 11 (B) 12

(C) 13 (D) 14

B

10. The number of ways of selecting two numbers from the set {1, 2……12} whose sum is divisible by 3 is

(A) 66 (B) 16

(C) 6 (D) 22

D

11. The number of 4-digit number that can be made with the digits 1, 2, 3, 4, 5 in which at least two digits are identical, is

(A) 45 - 5! (B) 600

(C) 54 - 4! (D) none of these

A

12. Number of triangles that can be formed joining the angular points of decagon is

(A) 30 (B) 20

(C) 90 (D) 120

D

13.(a2−a)C2 =

(a2−a )C4 , then a =



(A) 2 (B) 3

(C) 4 (D) none of these

B

14. The product of r consecutive integers is divisible by

(A) r (B) ∑k=1

r−1

k

(C) r ! (D) none of these

A,B,C

15. Total numbers of 5-digit telephone numbers that can be composed with distinct digits is

(A) 10P2 (B) 10P5

(C) 10C2 (D) none of these

D

16. If 7 points out of 12 are in same straight line, then number of triangles formed is

(A) 19 (B) 158

(C) 185 (D) 201

C

17. The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The total number of triangles that can be constructed by using these points as vertices is

(A) 220 (B) 204

(C) 205 (D) 195

C

18. How many numbers between 5000 and 10,000 can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 when each digit is appearing not more than once in each number?

(A) 5´8P3 (B) 5´8C3

(C) 5!´8P3 (D) 5!´8C3

A

19. The number of all the odd divisor of 3600 is

(A) 45 (B) 4



(C) 18 (D) 9

D

20. The number of 5-digit numbers in which no two consecutive digits are identical, is

(A) 92 ´ 83 (B) 9 ´ 84

(C) 95 (D) none of these.

C

21. The number of even proper divisors of 1008 is

(A) 23 (B) 21


22. The total number of ways in which a begger can be given at least one rupee from four 25-paise coins, three 50 paise coins and 2 one-rupee coins is (A) 54 (B) 50


23. The number of positive integers with the property that they can be expressed as the sum of the cubes of 2 positive integers in two different ways is

(A) 10 (B) 100

(C) 40 (D) ¥.

24. In an examination, the maximum marks for each of three papers are 50 each. Maximum marks for the fourth paper are 100. The number of ways in which the candidate can score 60% marks in the aggregate is

(A) 110556 (B) 110500


25. India and Sri Lanka play one day international seties until one team wins 4 matches. No match ends in a draw. The number of ways in which India can win the series is

(A) 35 (B) 70




26. The number of rectangles excluding squares from a rectangle of size 9 × 6 is

(A) 789 (B) 790


27. Eleven scientists are working on a secret project. They wish to lock up the documents in a cabinet such that the cabinet can be opened if and only if six or more scientists are present; then the least number of locks needed is

(A) 1 (B) 11 (C) 252 (D) 462

28. All the students of a class send New year greetings to one another. If the postman delivers 1190 greeting cards to the students of the class. Then the number of students in it is (A) 15 (B) 25

(C) 35 (D) 45

29. The number of permutations of 4 letter word taken from the word EXAMINATION having two alike and two different letters is

(A) 752 (B) 756

(C) 758 (D) 780

30. The total number of words that can be formed of the word ‘MATHEMATICS’ that begin with T and also end with T is

(A) 90720 (B) 907200

(C) 10080 (C) 91780

31. The number of ways of distributing 30 balls among three persons so that one person to get 10 balls , another persons 15 and 5.

(A)

30 !10 ! 15 ! 5! (B)

30 !10 ! 15 ! 5! .3!

(C)

30 !10 ! 15 ! 5! .3 (D) None of these

C

32. The number of proper divisors of 1008 is



(a) 23 (b) 28

(c) 20 (d) 24

33. In an examination there are three multiple choice questions (one option correct only) and each question has 4 choices. Number of ways in ways in which a student can fail to get all answer correct is

(A) 64 (B) 63

(C) 65 (D) None of these

34. Let Tn denotes the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1 – Tn = 21, then n equals

(a) 5 (b) 7

(c) 6 (d) 4

35. The number of words which can be made out of the letter of the word “MOBILE” when consonants always occupy odd places is

(A) 20 (B) 36

(B) 30 (D) 720

36. If 5 parallel straight lines are intersected by 4 parallel straight line, then the number of parallelogram formed is

(A) 60 (B) 207

(C) 101 (D) 126

37. The figures 4 , 5 , 6 , 7, 8 are written in every possible order. The number of numbers greater than 56000 is

(A) 72 (B) 90

(C) 96 (D) 98

38. The no. of arrangements of the letters of the word “BANANA” in which the two N’s do not appear adjacently is (a) 40 (b) 60(c) 80 (d) 100



39. The number of ways of selecting 2 white squares on a normal chess board, such that they don’t lie on the same row or column is equal to

(A) 96 (B) 400

(C) 480 (D) 496

40. Number of ways in which 6 persons can be seated around a table so that two particular persons are never seated together is equal to (a) 144 (b) 72(c) 36 (d) 240

41. Two players P1 and P2 plays a series of ‘2n’ games. Each game can result in either a win or loss for P1, total numbers of ways in which P1 can win the series of these games is equal to

(A)

12 (22n – 2nCn) (B)

12 (22n – 2 2nCn)

(C)

12 (2n – 2nCn) (D)

12 (2n – 2 2nCn

)

42. All the students of a class send New Year greetings to one another. If the postman delivers 1190 greeting cards to the students of the class. Then the number of students in it is (A) 15 (B) 25

(C) 35 (D) 45

43. If the (n + 1) numbers a, b, c, d……be all different and each of them a prime number, then the number of different factors (other than 1) of am. b. c. d……..is

(A) (m + 1) 2n (B) (m + 1) 2n – 1

(C) (m + 1) 2n + 1 (D) none of these

44. Give that n is odd, the number of ways in which three numbers in A.P. can be selected from 1, 2, 3, ….., n is

(A)

(n+1 )2

4 (B)

(n−1)2

4



(C)

(n+1 )2

2 (D)

(n−1)2

2

45. The number of rectangles that you can find on a chess board is

(A) 144 (B) 1296


46. The letters of the world “longest” are arranged at random. The probability that the vowels may occupy only odd positions, is

(A) 2/7 (B) 5/7

(C) 1/7 (D) 3/7

A

47. The number of arrangement of the letters of the word “Pencil” so that e always comes earlier then i , is

(A) 290 (B) 340

(C) 360 (D) 320

C

48. The no. of words which can be made from the letters of the word ‘BANANA’ is

(A) 60 (B) 120

(C) 360 (D) 720

A

49. The number of permutations of all the letters of the word ‘EXCERCISES’ is(A)60480 (B) 30240(C)151200 (D) 56504

C

50. There are 10 books and 3 copies of each. The number of ways in which a selection can be made from them.

(A) 103 (B) 310

(C) 410 - 1 (D) 114 - 1

B



51. The number of straight lines that can be formed by joining 20 points of which 4 points are collinear is

(A) 183 (B) 185

(C) 186 (D) 190

52. Number of ways in which 6 persons can be seated around a table so that two particular persons are never seated together is equal to (a) 144 (b) 72(c) 36 (d) 240

53. The number of ways in which 5 women can draw water from 5 taps, no tap remaining unused is

(A) 52 (B) 5!


54. The number of all the odd divisors of 3600 is

(A) 45 (B) 4

(C) 18 (D) 9

55. The number of 5 digit numbers that can be formed by using 1 and 2 is (A) 5! (B) 25 (C) 5C2 (D) none of these

56. The maximum number of points into which 4 circles and 4 straight lines intersect is

(A) 26 (B) 50

(C) 56 (D) 72

B

57. In how many ways the number 10800 be resolved as a product of two factors - (A) 35 (B) 36(C) 29 (D) 30

D

58. Number of ways in which the letters of the word ‘MATHEMATICS’ can be arranged without changing the order of vowels is



(A)

(11)!4 ! (2! )3

(B)

4 ! 7 !

(2 ! )3

(C)

411C . 7 !

(2 ! )2(D)

11!

(2 !)3

C

59. Number of nine digit numbers, which can be written down using the first nine natural numbers without repetition, If the digits 2 & 5 are never side by side is (A) 2(8!) (B) 8.(8!)(C) 7. 8! (D) 6. 8!

C

60. The results of 11 chess matches (as, win, lose or draw) are to be forcast. Out of all possible forcasts, the number of ways in which 8 correct & 3 incorrect results can be forcast is :

(A) 8! ´ 3! (B) 311C×81

(C) 811C×8 (D) none

C

61. The number of ways in which 5 beads, chosen from 8 different beads be threaded on to the ring is (A) 672 (B) 1344(C) 336 (D) none

A

62. The total of words that can be formed out of the word ‘MATHEMATICS’ that begin with A and also end with A is

(A)

11!

(2 !)3 (B)

11!

(2 !)2

(C)

9!

(2 !)2 (D)

9!

(2 !)2.

C

63. The number of ways in which Rs 20 can be distributed among five persons 50 that each person receives at least Rs 3 is

(A) 63 (B) 126


B



64. A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4, 5 without repetition total number of ways in which the number can be formed is

(A) 216 (B) 240

(C) 600 (D) 3125.

A

65. Six boys and six girls sit along a line alternately in x ways; and along a circle (again alternately) in y ways, then

(A) y = 10 x (B) y = 12 x

(C) x = 10y (D) x = 12y.

D

66. The number of ways in which three distinct numbering A.P can be selected from the set{1, 2, 3 …….12} is equal to

(A) 18 (B) 24

(C) 30 (D) 15.

C

67. The number of triangle whose vertices are at the vertices of an octagon but none of whose side happen to come from the side of the octagon is

(A) 16 (B) 15

(C) 54 (D) 27.

A

68. The total number of words that can be formed of the word ‘MATHEMATICS’ that begin with T and also end with T is

(A) 90720 (B) 907200

(C) 10080 (D) 91780

A

69. In an examination a minimum is to be secured in each of 5 subjects for a pass. The number of ways in which a student can fail is

(A) 31 (B) 36


A



70. The number of ways in which N positive signs and n negative sign (N³n) may be placed in a row so that no two negative signs are together is

(A) NCn (B) N+1Cn

(C) N! (D) N+1Pn

B

71. The number of ways of selecting 8 letters from 24 letters of which 8 are a, 8 are ‘b’ and the rest unlike is given by

(A) 27 (B) 8´28

(C) 10 ´27 (D) None of these

C

72. Number of ways in which 6 persons can be seated around a circular table so that two particular persons are never seated together is equal to

(A) 480 (B) 72

(C) 120 (D) 240

B

73. The number of diagonals of hexagon is

(A) 3 (B) 6

(C) 9 (D) 12

C

74. The number of 10 digits that can be written by using the digits 1 and 2 is

(A) 10´10 (B) 10P2

(C) 210 (D) 10!

C


(A) 45 (B) 4

(C) 18 (D) 9

D



LEVEL - II

1. The number of ways in which N positive signs and n negative signs (N ³ n) may be placed in a row so that no two negative signs are together is

(A) NCn (B) N+1Cn

(C) N! (D) N+1Pn

B

2. m men and n women to be sited in a row so that no two women sit together if m > n, then the number of ways in which they can be sited is

(A)

m ! n !(m+n ) ! (B)

m ! (m+1 ) !(m−n+1 ) !

(C)

m ! n !(m−n+1 ) ! (D) none of these

B

3. A parallelogram cut by n two sets of m lines parallel to its sides. The number of parallelogram thus formed is

(A) (mC2)2 (B) (m + 1C2)2

(C) (m + 2C2)2 (D) none of these

C

4. Number of permutations of n different objects taken r (³ 3) at a time which includes 3 particular objects, is

(A) nPr ´ 3! (B) nPr-3 ´ 3!

(C) n-3Pr-3 (D) rP3 ´ n-3Pr-3

D

5. The number of words from the letters of the word ‘BHARAT’ in which B and H never come together is

(A) 360 (B) 240


B

6. nCr-3 + 3nCr-2 + 3nCr-1 + nCr is equal to

(A) n+2Cr-1 (B) n+2Cr



(C) n+2Cr+1 (D) n+3Cr

D

7. Number of ways 6 different flowers can be given to 10 girls, if each can receive any number of flowers is

(A) 610 (B) 106

(C) 60 (D) 10C6

B

8.

14Cn

= 15Cn

+ 16Cn , then the value of n is

(A) 3 (B) 5

(C) 2 (D) 15

C

9. Number of ways in which 5 boys and 5 girls can form a circle such that boys and girls alternate is

(A) 2180 (B) 55

(C) 2880 (D) 2675

C

10. The least positive integral value of x which satisfy the inequality 10Cx –1 > 2.10Cx is

(A) 5 (B) 6

(C) 7 (D) 8

D

11. The number of ways in which 5 colour beads can be used to form a necklace, is

(A) 15 (B) 20

(C) 12 (D) 10

C

12. The number of different signals which can be given from 6 flags of different colours taking one or more at a time

(A) 1958 (B) 1956

(C) 16 (D) 64

B



13. Sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, 6 taken all at a time is

(A) 432 (B) 108

(C) 36 (D) 18

B

14. Number of ways in which 9 identical balls can be placed in three identical boxes is

(A) 55 (B)

9 !

(3 !)4

(C)

9 !

(3 !)3 (D) 12

D

15. Number of words that can be formed out of the letters of the word ‘COMMITTEE’ is

(A)

9 !

(2 ! )3 (B)

9 !

(2 ! )2

(C) (D) 9!

A

16. Number of six letters words that can be formed using the letters of word ‘ASSIST’ in which S’s alternate with other letter is

(A) 12 (B) 24


B

17. The number of ways in which ten candidate A1, A2, ….., A10 can be ranked such that A1 is always above A10, is

(A) 10! (B)

10 !2

(C) 2 ´ 10! (D) none of these

B

18. The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100 is

(A) 5081 (B) 6005




C

19. The number of zeros at the end of (127)! is

(A) 31 (B) 30

(C) 0 (D) 10

A

20. The number of permutations of the letters a, b, c, d, e, f, g such that neither the pattern ‘beg’ nor the pattern ‘cad’ appears is

(A) 4106 (B) 4806

(C) 4776 (D) 5120

B21. The number of ways in which 5 woman can draw water from 5 taps , no tap

remaining unused is

(A) 52 (B) 5!


B

22. In a college of 300 students , every student reads 5 newspaper and every newspaper is read by 60 students. The number of newspaper is

(A) less than 25 (B) at lost 20

(C) exactly 20 (D) exactly 25

D

23. Number of triangles which can be formed from 12 points out of which 7 are collinear

(A) 105 (B) 210

(C) 175 (D) 185

24. The number of n letter words that can be formed using letters of the word ‘AJMER’ such that it contain ‘A’ and ‘E’ & doesn’t contain ‘R’. (A) 4n + 2.3n – 2n (B) 4n + 2.3n + 2n (C) 4n – 2.3n + 2n (D) 4n – 2.3n – 2n

C

25. The number of different words of three letters which can be formed from the word “PROPOSAL”, if a vowel is always in the middle are



(A) 53 (B) 52(C) 63 (D) 32

26. The number of ways , in which 10 candidate A1, A2…..A10 can be ranked so that A1 is always above A2 is

(A) 10! (B) 10!/2

(C) 2.10! (D) 9!

27. The number of numbers, that can be formed by using digits 1,2,3,4,3,2,1 , so that odd digits always occupy odd places, is

(A) 3!4! (B) 34

(C) 18 (D) 12

28. The number of positive integral solution of abc = 30 is

(A) 25 (B) 26

(C) 27 (D) 28

29. Total number of divisors of n = 35 . 57. 79 that are of the form 4l + 1, l ³ 0, is equal to

(A) 240 (B) 30

(C) 120 (D) 15.

30. Number of even proper divisors of 1260 are

(A) 24 (B) 23

(C) 36 (D) 4.

31. The number of proper divisors of 2p . 6q . 15r is

(A) (p + q + 1)(q + r + 1)(r + 1) (B) (p + q + 1)(q + r + 1)(r + 1) – 2

(C) (p + q) (q + r) r – 2 (D) none of these

32. The number of odd proper divisors of 3p . 6m. .21n is



(A) (p + 1)(m + 1)(n + 1) – 2 (B) (p + m + n + 1)(n + 1) – 1

(C) (p + 1) (m + 1) (n + 1) – 1 (D) none of these.


(A) 23 (B) 21


34. The total number of ways in which a begger can be given at least one rupee from four 25-paise coins, three 50 paise coins and 2 one-rupee coins is

(A) 54 (B) 50


35. Number of triangles which can be formed from 12 points out of which 7 are collinear

(A) 105 (B) 210

(C) 175 (D) 185

36. In a room there are 12 bulbs of the same voltage, each having a separate switch. The number of ways to light the room with different amount of illumination.

(A) 122 – 1 (B) 212

(C) 212 – 1 (D) none of these

37. In an examination a candidate is required to pass four different subjects. The number of ways be can fail is

(A) 4 (B) 10

(B) 15 (D) 24

B

38. The number of six digit numbers, all digits of which are odd is (A) 54 (B) 56 (C) 55 (D) none of thee

B



39. The number of ways in which the digits of the number 125453752 can be rearranged such that no two 5’s come together is

(A)

9 !3! 2 ! (B)

7 !3! 2 !

(C)

37C . 6 !

2 ! (D) None

C

40. The number of ways in which 5 beads, chosen from 8 different beads be threaded on to the ring is (A) 672 (B) 1344(C) 336 (D) none

A

41. The number of ways in which Rs 20 can be distributed among five persons 50 that each person receives at least Rs 3 is

(A) 63 (B) 126

(C) 252 (D) 360

B

42. The number of words that can be formed from the letter a, b, c , d , e, f taken 3 at a times , each word containing at least one vowel is

(A) 96 (B) 84


A

43. There are three copies each of 4 different books, the number of ways in which they can be arranged in a self is

(A)

12 !

(3 ! )4(B)

11!

(3 ! )2

(C)

9 !

(3 ! )2(D) None of these

A

44. The number of lines drawn through 6 points lying on a circle is



(A) 12 (B) 15

(C) 24 (D) 30

B

45. 12 persons are to be arranged to a round table if two particular persons among them are not to be side by side, the total number of arrangement is

(A) 9(10)! (B) 2(10)!

(C) 45(8)! (D) 10!

B

46. Four boys picked 30 apples , the number of ways in which they can divide then if all the apples are identical is

(A) 5630 (B) 4260


C

47. n and m are selected from {1, 2, 3, 4, …, 20} with replacement. Number of ways selecting (m, n) such that 7m + 7n is divisible by 10, is

(A) 25 (B) 100

(C) 625 (D) 150

B

48. Number of ways in which a 10 digit number can be formed using {1, 2, 3, 4, 5, - 6}, such that sum of 1st and 2nd digit, 3rd and 4th digit, …, so on is 10.

(A) 25 (B) 35

(C) 45 (D) 55

B

49. A double decker bus can accommodate (u + l) passengers, u in the upper deck and l in the lower deck. The number of ways in which the (u + l) passengers can be distributed in the two decks if r(£l) perticular passengers refuse to go in the upper deck and s(£u) refuse to sit in the lower deck is

(A) u-sCl - r (B)

(u+l−r−s )!( l−r )! (u−s )!



(C) u - sPl – r (D)

(u+l)!r ! s !

B

50. The number of integral solutions of x + y + z = 0 with x, y, z ³ -4 is (A) 15C7 (B) 14C12

(C) 17C15 (D) 17C10

B

51. The number of 4 digit numbers having non-zero digits such that sum of the digits is 10 is (repetition of digits not allowed)

(A) 42 (B) 34

(C) 84 (D) 24

D

52. The number of ways of switching the network such that the bulb glows is

(A) 61 (B) 6


A


(A) 23 (B) 21

(C) 20 (D) 24

D

54. The number of non-negative integral solutions of x1 + x2 + x3 + 4x4 = 20 is

(A) 520 (B) 536


B

55. The number of permutations of 4 letter word taken from the word EXAMINATION having two alike and two different letters is (A) 752 (B) 756



(C) 758 (D) 780

B

56. Number of ways in which two smaller square can be selected from a chessboard such that they have exactly one corner common

(A) 98 (B) 216

(C) 112 (D) 90

D

57. Number of ways of distributing 10 identical objects among 8 persons(one or many persons may not be getting any number of objects) is

(A) 810 (B) 108

(C) 717C (D) 8

10C

C

58. In how many ways the number 10800 be resolved as a product of two factors -(A) 35 (B) 36(C) 29 (D) 30

D

59. The number of ways in which three distinct numbering A.P can be selected from the set{1, 2, 3 …….12} is equal to

(A) 18 (B) 24

(C) 30 (D) 15.

C

60. All the students of a class send New year greetings to one another. If the postman delivers 1190 greeting cards to the students of the class. Then the number of students in it is (A) 15 (B) 25(C) 35 (D) 45

61. The number of different nine digit number that can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions are(A) 16 (B) 36(C) 60 180




(A) 45 (B) 4(C) 18 (D) 9

63. If a, b, c, d, e are prime integers, then the number of divisors of ab 2c2de excluding 1 as a factor, is

(A) 94 (B) 72(C) 36 (D) 71

64. In a one day cricket match between India and West Indies, the Indians require 10 runs in the last 3 balls to win. If any one of the scores 0, 1, 2, 3, 4, 5, 6 can be made from a ball and no wides or no – balls are bowled, then the number of different ways that batsmen make exactly 10 runs is

(A) 15 (B) 21


65. The number of words of 6 letters that can be formed with the letters of the word ABRAKADABRA, if the word begins and end with A are

(A 270 (B) 290

(C) 370 (D) 345

66. A polygon has 44 diagonals, then n is equal to (A) 10 (B) 11(C) 12 (D) 13

B

67. nCr + 2.nCr + 1 + nCr+2 is equal to (2 £ r £ n)(A) 2.nCr + 2 (B) n +1Cr +1

(C) n + 2Cr + 2 (D) none of theseC

68. The number of five digits telephone numbers having atleast one of their digits repeated is (A) 90,000 (B) 100000(C) 30240 (D) 69760

D



69. How many words can be made from the letters of the word INSURANCE, if all vowels come together

(A) 18270 (B) 17280


D

70. If a, b, c, d, e are prime integers, then the number of divisors of ab 2c2de excluding 1 as a factor, is

(A) 94 (B) 72

(C) 36 (D) 71

D

71. Number of all four digit numbers having different digits formed of the digits 1, 2, 3, 4 and 5 and divisible by 4 is

(A) 24 (B) 30

(C) 125 (D) 100

A

72. Let A be the set of 4-digit numbers a1a2a3a4 where a1> a2> a3> a4, then n(A) is equal to

(A) 126 (B) 84


C

73. Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A to B is

(A) 144 (B) 12

(C) 24 (D) 64

C

74. If A and B are disjoint finite sets then number of elements in A È B is



(A) n(A Ç B) (B) n(A –B)

(C) n(A) ´ n (B) (D) n(A) + n (B)

D

75. If x = {1, 2, 3, 4, 5, 6, 7}, then the number of proper subsets of x, containing 7, is

(A) 60 (B) 61

(C) 62 (D) 63

D



ANSWERS (Objective)

LEVEL– I

1. B 2. B

3. A 4. D

5. C 6. A

7. A 8. B

9. B 10. D

11. A 12. D

13. B 14. A, B, C

15. D 16. C

17. C 18. A

19. D 20. C

LEVEL -II

1. B 2. B

3. C 4. D

5. B 6. D

7. B 8. C

9. C 10. D

11. C 12. B

13. B 14. D

15. A 16. B

17. B 18. C

19. A 20. B


permutation and combination

Documents

art of counting combinatorics

possible combinations

different codes

art of counting introduction

possible number

possible cases

addition principle

multiplication principle