chapter 24 permutation and combination

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Chapter 24

PERMUTATION AND COMBINATION

Factorial Notation

The factorial n is the continued product of n consecutive natural numbers

beginning with 1 is denoted by n!

Therefore,

n! = 1x 2 x 3 x 4 x 5 x………….(n – 2)x (n – 1)x n

Thus we have for example

6! = 1 X 2 X 3 X 4 X 5 X 6 = 720

5! = 1 X 2 X 3 X 4 X 5 = 120

4! = 1 X 2 X 3 X 4 = 24 and so on

1. Fundamental Principle of Multiplication: If there are two jobs such

that one of them can be completed in m ways, and when it has been

completed in any one of these m ways, second job can be completed in n

ways, then the two jobs in succession can be completed in m x n ways.

For example, suppose there are 12 boys and 9 girls in a class. The class

teacher wants to select 1 boy and 1 girl to represent the class in a Quiz

Competition. In how many ways can the teacher make this selection?

The teacher can perform the two selections as follows

Selecting a boy from 12 boys

Selecting a girl from 9 girls

The first job can be done in 12 ways and the second can be done in 9

ways.

According the fundamental Principle of Counting, the total number of ways

this can be done is




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12 x 9 = 108 ways

2. Fundamental Principle of Addition: If there are two jobs such that they

can be preformed independently in m and n ways respectively, then either

of the two jobs can be performed in (m + n) ways

Consider the above mentioned example again,

If the teacher want to select either a boy or a girl to represent the class in

the Quiz Competition, This can be done as follows

Selecting a boy = 12 ways

Selecting a girl = 9 ways

Thus the total number of ways = 12 + 9 = 21 ways

Arrangement of different objects

Suppose we have four objects P, Q R and S. We have to form a group of

two objects out of these given four objects. Thus we are actually forming

groups of four different objects taking two at a time. Thus we have the

following six groups;

(i) PQ

(ii) PR

(iii) PS

(iv) QR

(v) QS

(vi) RS

But the objects in each of these groups can be arranged in two different

ways. Take the first group. This group can be arranged as PQ or QP. And

similarly other groups also. Thus we a get a total of 12 such arrangements.

Effectively, we have




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Total Number of Arrangements = The total Number of groups X r! , where r

is the number of objects in each group.

Thus, total number of arrangements = 6 x 2! = 12

Here the first case of number of groups denote the Combinations where the

order of the arrangements of the objects does not matter, where as the

second case of number of arrangements represent the Permutations,

where the order of arrangements of the objects are important.

Definition of Combination

Each of the different selections or groups which can be made by selecting

some or all of a number of given objects at a time.

The number of Combinations of different objects taken r at a time can be

represented by the symbol, nCr

nCr = 𝑛!

𝑟!( 𝑛−𝑟)!

For example, the symbol 10C3 denotes the number of selections of ten

different objects taken three at a time, and can be represented as

10C3 = 10!

3!(10− 3)! =

10!

3! 7! =

10×9×8×7×6×5×4×3×2×1

3×2×1 ×7×6×5×4×3×2×1 = 120

Definition of Permutation:

Each of the different arrangements which can be made by taking some or

all of the given things or objects at a time is called a Permutation. The

number of Permutation of n different objects taken r at a time can be

represented by the symbol nPr, where the letter P stands for Permutation.

Thus we have

nPr = 𝑛!

(𝑛−𝑟)!




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For example, the symbol 10P3 denotes the number of permutations of 10

different objects taken 3 at a time.

Thus we have

10P3 = 10!

(10−3)! =

10×9×8×7×6×5×4×3×2×1

7×6×5×4×3×2×1 = 720

We observe that, 10P3 = 10C3 X 3!

Thus we have

nPr = nCr X r!

We have

nPr = 𝑛!

(𝑛−𝑟)!

Put r = n

nPn = 𝑛!

0!

n! = 𝑛!

0!

⇒ 0! = 1 and

nCn = nC0 = 1

Difference between a Permutation and a Combination:

1. In a combination only selection is made where as in a permutation not

only a selection but also an arrangement in a definite order is very

important.

2. Generally the word ‘arrangements’ are used for Permutations and the

word ‘selection’ is used for Combinations

3. Each combination correspond to many Permutations

More Examples:




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Example 1. How many quadrilaterals can be formed by joining the vertices

of an octagon?

We know that a quadrilaterals has four sides and four vertices n where as

an octagon has 8 sides and 8 vertices

Therefore, the required number of quadrilaterals

8C4 = 8!

4!(8−4)! =

8×7×6×5

24 = 70

Example 2. In how many ways three different rings can be worn in four

fingers with at most one in each finger?

It is the same as the number of arrangements of 4 fingers, taken 3 at a time

Therefore, the required number of ways 4P3 = 4!

(4−3)! = 24

Example 3 In how many ways can 6 persons stand in a queue?

It is same as the number of arrangements of 6 different things taken all at a

time.

Thus, the required number of ways = 6P6 = 6! = 720

Example 4. How many triangles can be formed by joining the vertices of a

hexagon?

For a hexagon there are six vertices, whereas for a triangle there are only

three vertices.

Therefore , the required number of triangles = 6C3 =20

Example 5 In a party every person shakes hands with every other person.

If there was total of 210 handshakes in the party, find the number of

persons who were present in the party.

Let n be the total number of persons in the party. For each selection of two

persons there will be a handshake.




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Therefore, number of handshakes = nC2 = 210

𝑛 𝑋 (𝑛 – 1)

2 = 210

n x (n – 1) = 2 x 210 = 21 x 20

Therefore, n = 21

EXERCISE

(Questions based on Permutation)

1. In how many different ways can the letters of the word CREAM be

arranged?

(a) 120 (b) 480 (c) 240 (d) 125 (e) None of these

2. In how many ways 7 persons can be arranged in a circle?


Note: In a circular permutation, there is no first or last place of an object.

Therefore the principles of linear permutations are not possible. In

circular permutations, the relative positions of the things alone need to be

taken into consideration and not actual positions.

Therefore, No of circular Permutations of n different things taken all at a

time round a circle is (𝑛 − 1)!

3. In how many different ways can the letters of the word CREATE be

arranged?


Note: Permutation of repeated objects: The number of distinct

permutations of n things taken all at a time, when p of them are similar

and of one type and q of them are similar and of another type, and r of

them are similar and of third type is : 𝑛!

𝑝!𝑞!𝑟!

4. How many three letter words can be made using the letters of the

word ORIENTAL ?





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5. How many words can be formed from the letters of the word

DIRECTOR so that the vowels always come together?


6. In how many different ways can the letters of the word RANDOM

be arranged?


7. In a shelf there is space for 4 books. In how many ways can it be

filled with 12 different books?


8. Four boys and three girls are to be seated in a row for a photograph

such that two girls are always together. In how many different ways

can they be seated?


9. In how many different ways can the letters of the word AROUND be

arranged?


10. Five members of a board are to be seated for a meeting. Five chairs

are put in the meeting room for them. In how many ways can they

take their seat?


(Questions based on Combination)

11. If there are 12 members in a room and if each two of them shake

hands with each other, how many handshakes happen in the room?


Directions(Q 12 – 13): A team of four members is to be selected from a

group of 4 boys and 3 girls:

12. In how many different ways the team can be formed so that there will

be equal number of boys and girls?





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13. In how many different ways it can be done so that there will be at

least one girl in the team?

(a) 46 (b) 35 (c) 34 (d)24 (e) None of these

14. In an office there are 20 males and10 females. In how many ways the

manager can select either a man of a woman to represent the office?


15. A bag contains 10 blue balls and 6 Green balls. Find the way of

selecting 2 Blue and 3 green balls out of them.

(a)65 (b) 90 (c) 900 (d)9000 (e) None of these

16. In how many ways a team consisting of 5men and 6 women can be

formed from 8men and 10 women?

(a)266 (b)86400 (c)11760 (d)5040 (e)None of these

17. In a class there are 30 boys and 22 girls. There are to be divided in to

subgroups each containing either 2 boys or 3 girls. What is the

maximum number of ways of forming such groups?


18. In a cricket tournament, every team played one match with the other

team. What is the total number of matches played if five teams

participated in the tournament?


Directions (Qs. 19 to 20): A group of 5 members is to be formed by

selecting out of 4 boys and 5 girls

19. In how many different ways the groups can be formed if it should

have 2 boys and 3 girls


20. In how many different ways the group can be formed if it should have

at least 1boy?





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chapter 24 permutation and combination

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