Principles of Counting and Theories of Probability

In some banks, depositors can deposit or withdraw money anytime through an automated teller. They just insert a card containing some information into an automated teller machine, press their four-digit number code in the keyboard. The computerized machine reads the code in the card, and then it processes the transactions bases on the input of the depositors. Can you imagine now, how many different number codes can be made using four digit at a time? A problem like this needs knowledge on counting techniques.

The Fundamental Counting Principle

If activity 1 can be done in n1 ways, activity 2 can be done in n2 ways, activity 3 can be done in n3 ways, and so forth; then the number of ways of doing these activities on a specified order is the product of n1,n2,n3 and so on. In symbols,

n1 x n2 x n3… nn

Example 1:Suppose a school has three gates, in how many ways can a student enter and leave the

school?

Solution:There are two activities involved here, the first is entering and the second is leaving the

school. The student has three choices in doing the first activity and so on with the second activity. Suppose we call the first gate as A, the second as B, and the third as C, let us explore some possible ways of solving the problem.

1. By listingThe different entrance-exit pairs are: AA, AB, AC, BA, BB, BC, CA, CB, CC

There are nine different ways of doing the said activities

2. By using a table

Entrance Gate

Exit Gate A B C

A (A,A) (A,B) (A,C) B (B,A) (B,B) (B,C) C (C,A) (C,B) (C,C)

In the table, the ordered pair (A, A) means that the student can enter and leave the school using the same gate. The order pair (A, B) means that the student can enter the school in gate A and leave using gate B.

3. By using a tree diagram

Entrance Exit Way of doing Gate Gate the Activities

A A A A B A B

C A C

A B A START B B B B

C B C

A C A C B C B

C C C

The diagram shown above shows the nine different ways of entering and leaving the school.

4. By using the fundamentals counting principle

There are three gates to enter the school, n1 =3; and using the same for leaving; n2 =3 thus, 3 x 3 = 9 different ways.

The Counting Principle for Alternative Cases

Suppose the ways of doing an activity can be broken down into several alternative cases where each cases does not have anything in common with the other cases. If case 1 can be done in n1 ways, case 2 can be done in n2 ways, case 3 can be done in n3 ways, and so on, then the number of ways the activity can be done is the sum of n1, n2, n3 and so on cases. In symbols,

n1 x n2 x n3… + nn

Example:How many different fractions less than 1 can be formed using the numbers 2, 3,

5, 7, and 9?

Solution:The activity involved here is forming fractions that are less than 1 from the given

numbers above. For a fraction to be less than 1, the numerator should be the less than the denominator. The number to be used as numerator should be taken into consideration when forming the fractions. To solve this, different cases have to be considered.

Case 1: When the numerator is 2, there are 4 possible denominators that can be used. These are 3, 5, 7, and 9

Case 2: When the numerator is 3, there are 3 possible denominators that can be used. These are 5, 7, and 9.

Case 3: When the numerator is 5, there are 2 possible denominators that can be used. These are 7, and 9.

Case 4: When the numerator is 7, only 9 is the number that can be used as denominator

Case 5: When the numerator is 9: No more number can be used as denominator

Therefore, 10 different fractions less than 1 can be formed from the digits 2, 3, 5, 7, and

9. These fractions are 23

, 25

, 27

, 29

, 35,37

, 39,59,∧79

.

Factorial (n!)

Factorial n denoted by n!, and is defined asn! = n(n – 1)(n – 3) … 1

The product 3 x 2 x 1 can be written in brief factorial notation as 3! (read as “three factorial” or “factorial 3”). Thus, seven factorial is written as:

7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040

Other examples:4! = 4 x 3 x 2 x 1 = 245! = 5 x 4 x 3 x 2 x 1 = 1200! = 1

Permutation

The term permutation refers to the arrangement of objects with reference to order. Given a set with n objects, then we can take r objects from the set. The total number of n objects taken r at a time is represented by the notation ❑nPr and can be evaluated using the formula,

P(n, r) = n !

(n , r )

Where n is the number of objects and r is the number of objects taken from n at a time.

The notation ❑nPr or P(n,r) is read as “the number of permutations of n objects taken r at a time”.

Permutation Rule 1

The number of permutations of n distinct objects taken all together is n!.

Example 1:How many different signals can be made using five flags if all the flags must be used in

signals?

Solution:P(5, 5) = 5! = 120 different signals

Example 2:In how many ways can 5 people line up for a group picture ifa) Two want to stand next to each otherb) Two refuse to stand next to each other

Solution:a) Since 2 people want to be together, we can consider them as 1 person. The original 5

people can be considered as 4 people. We have to consider 2 activities here:

Permuting the 4 people; and Permuting the 2 people who want to stand together.

There are 4! or 24 ways of arranging the 4 people. The 2 people who want to stand together can be arranged in 2! ways. Therefore, using the FCP, there are 4! x 2! = 48 ways the 5 people can line up given in condition a.

b) The condition that two people refuse to stand next to each other is just the opposite of the condition that they want to stand next to each other. Therefore, we can get the answer to (b) by subtracting the answer in (a) from the total number of permutations of the 5 people. Which is 5!.

5! – 48 = 120 - 48 = 72 ways

Permutation Rule 2

The arrangement of n objects in order using r objects at a time is given by the formula

nPr = n !

(n−r ), where r < n.

Example 1:How many different ways can a chairperson and an assistant chairperson be selected for

a research project if there are seven statisticians available?

Solution:n = 7; r = 2

❑7P2 = 7 !

(7−2 )! = 7 !5!

= 42 ways

Example 2:Suppose these are eight machines, but only these spaces in the display room are

available for the machines. In how many different ways can the 8 machines be arranged in the three available spaces?

Solution:There are 8 machines: n = 8 but only 3 machines can be displayed r = 3.

❑8P3 = 8!

(8−3 )! = 8 !5!

= 8 x7 x6 x 5 !

5 ! = 336 ways

Permutation Rule 3

The arrangement of n objects in a circular pattern is given by the formula P = (n – 1)!.

Example 1:In how many ways can six persons be seated around a circular table?

Solution:Since this is a circular permutation of 6 things, there are (6 – 1)! or 120 possible seating

arrangements.

Example 2:In how many ways can 8 beads be put together to form a bracelet?

Solution:n = 8P = (n – 1) = (8 – 1)! = 7! = 5,040 ways

Permutation Rule 4

The number of permutations of n objects in which r1 are alike, r2 are alike, r3 are alike, … etc, is

P = n!

r1!r 2!r3 !…rn ! where r1 + r2 + r3 + rn = n.

Example 1:How many different permutations can be made from the letters of the word

“MISSISSIPPI”?

Solution:Regroup the letters as MIIIISSSSPP. The total number of letters is n = 11, where r1 = 1

(there is 1 M), r2 = 4 (there is 4 I’s), r3 = 4 (there are 4 S) and r 4 = 2 (there are 2 P’s).

P = n !

r1!r 2!r3 !…rn =

11!1! 4 !4 !2!

= 39,916,8001,152

= 34,650 different permutations Example 2:

How many distinguishable permutations of the letters in the word INTERFERENCE?

Solution:n = 12; r1 = 1 (I); r2 = 2 (N); r3 = 1 (T); r 4 = 2 (R); r5 = 1 (F); r6 = 1 (C); r7 = 4 (E).

P= n!

r1!r 2!r3 !…rn ! =

12 !1!2 !1 !2!1 !1! 4 !

= 479,001,600

96

= 4,989,600 ways

CombinationsSuppose we are interested by only In the number of different ways that r objects can be

selected from a given number of objects. If the order of the objects is not important, the total number of orders or arrangement is called combination.

The number of combinations of n objects taken r at a time is denoted by ❑nC r or C(n, r) and is given by the formula:

❑nC r = n !

(n−r )!r !

Notice that the formula for ❑nC r is the same in the permutation formula n !

(n−r ) with an

r! in the denominator. This r! divides out the duplicates from the number of permutations.

Example 1:Given the letters A, B, C, and D, list the permutations and combinations for selecting two

letters.

Solution:Using the formula,

❑nPr = n !

(n−r )! ❑4 P2 =

4 !(4−2 ) ! =

4 !2 !

= 12 different ways

❑nC r = n !

(n−r )!r ! = ❑4C2 = 4 !

(4−2 ) !2! = 4 !2!2 !

= 6 different ways

Permutations CombinationsAB BA CA DAAC BC CB DBAD BD CD DC

AB BCAC BDAD CD

Note that in permutation, AB is different from BA but in combination, AB is the same as BA.

Example 2:In order to survey the opinions of costumers at local malls, a researcher decides to

select 5 malls from a certain area with a total of 9 malls. How many different ways can be selection be made?

Solution:

❑9C4 = 9!

(9−5 )5 ! = 9!4 !5 !

= 126 different ways

Example 3:In a club there are 8 women and 5 men. A committee of 4 women and 2 men is to be

chosen. How many possibilities are there?

Solution:Here, one must select 4 women from 8 which can be done in❑8C4 or 70 ways. The next

activity is to choose 2 men from 5 men, which can be done in ❑5C2 or 10 ways. Finally, by the FCP, the total number of different ways is ❑8C4 x ❑5C2 = 70 x 10 = 700 ways, since one is choosing both men and women.

Example 4:A committee of 5 people must be selected from 5 accountants and 8 educators. How

many ways can the selection be done if there are at least 3 educators in the committee?

Solution:A committee can consist of 3 educators and 2 accountants, or 4 educators and 1

accountant, or 5 educators. To find the different possibilities, find each separately, and then add them.

❑8C3 . ❑5C2 + ❑8C4 . ❑5C1 + ❑8C5 = 8 !5!3 !

. 5 !3! 2!

+ 8 !3!5 !

= 56 . 10 + 70 . 5 + 56 = 560 +350 + 56

= 966 different committees

Note: the word and means to multiply, and the word or means to add.