2.7. Counting principle. Permutation and combination

Counting principle:

If the set E contains n elements and the set F contains m elements, there are ways in which we can choose first an element of E and then element

of F.

Example:

We toss a coin two times. This experiment has two steps: the first step toss, the second toss. Each step has two outcomes: a head and a tail. Thus, total outcomes for two tosses of a coin= .

The four outcomes for this experiment are: HH, HT, TH, TT

Generalized counting principle:

Let be sets with elements, respectively. Then there are ways in which we can first choose an element of , then an element of ,……., and finally an element of .

Example:

How many outcomes does the experiment of throwing five dice have?

Solution:

Let , be set of all possible outcomes of die. Then . The number of the outcomes of the experiment of throwing five dice equals the number of ways we can first choose an element of , then an element of ,….. , and finally an element of .

That is

2.7.1. Permutation

Definition:

An n-element permutation of a set with n objects is simply called a permutation, denoted by . The number of permutations of a set containing n elements is

Example:

The 3 digits 1, 2, 3 can be arranged in different orders:

123, 132, 213, 231, 312, 321

Definition:

The number of permutations, of r objects chosen from n is the number of possible arrangements when r objects are to be selected from a total of n and arranged in order. This number is

Remark:

Instead of , the symbols and P (n, r) are frequently used to denote the number of permutations of n objects taken r at a time. Different authors frequently use different symbols.

Example:

Three students, Kanat, Askhat, and Marat must be scheduled for a job interviews. In how many different orders can this be done?

Solution:

The number of different orders is equal to the number of permutations of the set {Kanat, Askhat, Marat}. So there are 3!=6 possible orders for the interviews.

Example:

If 5 persons are to pose for a photograph by standing in a row, how many different arrangements are possible?

Solution:

Since we are arranging 5 “objects” 5 at a time, the number of different arrangements is given by .

Example:

If five boys and five girls sit in a row in a random order, what is the probability that no two children of the same sex sit together?

Solution:

There are 10! ways for 10 persons to sit in a row. In order that no two of the same sex sit together, boys must occupy positions 1, 3, 5, 7, 9 and girls positions 2, 4, 6, 8, 10, or vice versa. In each case there are possibilities. So, the desired possibility is .

Theorem: The number of distinguishable permutations of n objects of k different types, where are alike, are alike,…… are alike and is

.

Example:

How many different 10- letter codes can be made using three a’s , four b’s , and three c’s?

Solution:

By theorem, the number of such codes is .

2.7.2. Combination

If order is of no importance, then we have a combination rather a permutation. A combination of n objects taken r at a time is a selection of r objects taken from the n, without regard to the order in which they are selected or arranged. Order is irrelevant.

Definition:

The number of combinations, , of r objects chosen from n is the number of possible selections that can be made. This number is

Remark:

Some other symbols which are used to denote the number of combinations of n objects taken r at a time are , C (n, r) and .

Example:

If a club has a membership of ten, then how many three-man committees are possible?

Solution:

Order is not important, so this is combination problem. The number of possible committees is equal to the number of ways three persons can be selected from ten persons, namely .

We have:

Example:

In how many ways can two math and three biology books be selected from eight math and six biology books?

Solution:

There are possible ways to select two math books and possible ways to select three biology books. Therefore, by counting principle, is the total number of ways in which two math and three biology books can be selected.

Example:

A box contains 24 transistors, four of which are defective. If four are sold at random, find the following probabilities:

a) Exactly two are defective b) None is defective

c) All are defective d) At least one is defective

Solution:

There are ways to sell four transistors, so the denominator in each case will be =10626

a) Two defective transistors can be selected as and two nondefective ones as . Hence

b) The number of ways to choose no defective is .

Hence

c) The number of ways to choose four defectives from four is , or 1. Hence

d) To find the probability of at least one defective transistor, find the probability that there are no defective transistors, and then subtract that probability from 1.

.

Exercises

1. How many permutations of the set begin with a and end

with c?

2. How many different messages can be sent by five dashes and three dots?

3. Roman has eight guests, two of whom are Jane and John. If the guests will arrive in a random order, what is the probability that John will not arrive right after Jane?

4. Find the number of distinguishable permutations of the letters MISSISSIPPI.

5. There are 20 chairs in a room numbered 1 through 20. If eight girls and 12 boys sit on these chairs at random, what is the probability that the thirteenth chair is occupied by a boy?

6. If we put five math, six biology, eight history, and three literature books on a bookshelf at random, what is the probability that all the math books are together?

7. Five boys and five girls sit in a row at random. What is the probability that the boys are together and the girls are together?

8. A man has 20 friends. If he decides to invite six of them to his birthday party, how many choices does he have?

9. A panel consists of 20 men and 20 women. How many choices do we have for a jury of six men and six women from this panel?

10. In a company there are seven executives: four women and three men. Three are selected to attend a management seminar. Find the following probabilities:

a) All three selected will be women.

b) All three selected will be men.

c) Two men and one woman will be selected.

d) One man and two women will be selected.

11. In a class of 18 students, there are 11 men and seven women. Four students are selected to present a demonstration on the use of the calculator. Find the probability that the group consists of the following:

a) All men

b) All women

c) Three men and one woman

d) One man and three women

e) Two men and two women.

12. A committee of four people is to be formed from six doctors and eight dentists. Find the probability that the committee will consist of the following:

a) All dentists

b) Two dentists and two doctors

c) All doctors

d) Three doctors and dentist

e) One doctor and three dentists

13. From a faculty of six professors, six associate professors, 10 assistant professors, and 12 instructors, a committee of size 6 is formed randomly. What is probability that

a) There are exactly two professors on the committee?

b) All committee members are of the same rank?

14. Almas has three sets of classics in literature, each set having four volumes. In how many ways can he put them in a bookshelf so that books of each set are not separated?

Answers

1. 6; 2. 40320; 3. 0.875; 4. 34 650; 5. 0.6; 6. 0.00068; 7. 0.0079; 8. 38 760;

9. 1502337600; 10. a) 4/35; b) 1/35; c) 12/35; d) 18/35; 11. a) 11/102;

b) 7/612; c) 77/204; d) 77/612; e) 77/204; 12. a) 10/143; b) 60/143;

c) 15/1001; d) 160/1001; e) 48/143; 13. a) 0.228; b) 0.00084; 14. 82944.