PERMUTATION AND COMBINATION
PROBLEMS

About "Permutation and Combination Problems"

Permutation and Combination Problems :

In this section, we are going to learn, how to solve problems on permutation and combination step by step.

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Permutation and Combination Problems

Problem 1 :

A committee of 7 members is to be chosen from 6 artists, 4 singers and 5 writers. In how many ways can this be done if in the committee there must be at least one member from each group and at least 3 artists ?

Solution :

For the given condition, possible ways to select members for a committee of 7 members.

(3A, 3S, 1W) ----> 6C3 ⋅ 4C3 ⋅ 5C1 = 20 ⋅ 4 ⋅ 5 = 400

(3A, 1S, 3W) ----> 6C3 ⋅ 4C1 ⋅ 3C1 = 20 ⋅ 4 ⋅ 10 = 800

(3A, 2S, 2W) ----> 6C3 ⋅ 4C2 ⋅ 5C2 = 20 ⋅ 6 ⋅ 10 = 1200

(4A, 2S, 1W) ----> 6C4 ⋅ 4C2 ⋅ 5C1 = 15 ⋅ 6 ⋅ 5 = 450

(4A, 1S, 2W) ----> 6C4 ⋅ 4C1 ⋅ 5C2 = 15 ⋅ 4 ⋅ 10 = 600

(5A, 1S, 1W) ----> 6C5 ⋅ 4C1 ⋅ 5C1 = 6 ⋅ 4 ⋅ 5 = 120

Thus, the total No.of ways is

= 400 + 800 + 1200 + 450 + 600 + 120

= 3570

Hence, the no. of ways,a committee of 7 members can be chosen is 3570.

Let us look at the next problem on "Permutation and combination problems".

Problem 2 :

The supreme court has given a 6 to 3 decisions upholding a lower court. Find the number of ways it can give a majority decision reversing the lower court.

Solution :

Upholding a lower court means, supporting it for its decision.

Reversing a lower court means, opposing it for its decision.

In total of 9 cases (6 + 3 = 9), it may give 5 or 6 or 7 or 8 or 9 decisions reversing the lower court. And it can not be 4 or less than 4. Because majority of 9 is 5 or more.

The possible combinations in which it can give a majority decision reversing the lower court are

5 out of 9 ----> 9C5 = 126

6 out of 9 ----> 9C6 = 84

7 out of 9 ----> 9C7 = 36

8 out of 9 ----> 9C8 = 9

9 out of 9 ----> 9C9 = 1

Thus, the total no. of ways is

= 126 + 84 + 36 + 9 + 1

= 256

Hence the no. of ways it can a majority of the decision reversing the lower court is 256.

Let us look at the next problem on "Permutation and combination problems".

Problem 3 :

Five bulbs of which three are defective are to be tried in two bulb points in a dark room. Find the number of trials in which the room can be lighted.

Solution :

Given : 3 bulbs are defective out of 5.There are two bulb points in the dark room

One bulb (or two bulbs) in good condition is enough to light the room.

Since there are two bulb points, we have to select 2 out of 5 bulbs.

No. of ways of selecting 2 bulbs out of 5 is

= 5P2

= 10

(It includes selecting two good bulbs, two defective bulbs, one good bulb and one defective bulb. So, in these 10 ways, room may be lighted or may not be lighted)

No. of ways of selecting 2 defective bulbs out of 3 is

= 3C2

= 3

(It includes selecting only two defective bulbs. So, in these 3 ways, room can not be lighted)

The number of ways in which the room can be lighted is

= 10 - 3

= 7

Hence the no. of trials in which the dark room can be lighted is 7.

Let us look at the next problem on "Permutation and combination problems".

Problem 4 :

Find the number of ways of selecting 4 letters from the word EXAMINATION.

Solution :

There are 11 letters in the word of which A,I,N are repeated twice.

Thus, we have 11 letters of 8 different kinds as given below.

(A, A), (I, I), (N, N), E, X, M, T, O

The group of 4 letters can be selected in any one of the following 4 forms.

(i) 2 alike and other 2 alike.

(ii) 2 alike and other 2 different.

(iii) All 4 different.

Case (i) :

If 2 are alike and other 2 are also alike, any 2 of the 3 groups

(A, A),(I, I),(N, N)

will be selected.

The number of ways is

= 3C2

= 3

Case (ii) :

If 2 are alike and other 2 are different, any one of the three groups

(A, A), (I, I), (N, N)

and 2 letters from 7 different letters are selected.

[E, X, M, T, O + 2 different letters from (A, A), (I, I),(N, N), because one of the groups is already selected]

The number of ways is

= 3C1 ⋅ 7C2

= 3 ⋅ 21

= 63

Case (iii) :

If all four are different, 4 from 8 different letters

A, I, N, E, X, M, T, O

are selected.

The number of ways is

= 8C4

= 70

Thus the total number of ways is

= 3 + 63 + 70

= 136

Hence, the number of ways of selecting 4 letters from the word EXAMINATION is 136.

Let us look at the next problem on "Permutation and combination problems".

Problem 5 :

The letters of the word ZENITH are written in all possible orders. If all the words are written in a dictionary, what is the rank or order of the word ZENITH ?

Solution :

No. of new words formed with the letters of the word

ZENITH = 6! = 720

Alphabetical order of the letters of the word ZENITH is

E,H,I,N,T,Z

Dictionary gives meanings in the order starting with E,H and so on.

Number of words can be formed starting with E,H and so on.

E __ __ __ __ __ = 5! = 120 words

H __ __ __ __ __ = 5! = 120 words

I __ __ __ __ __ = 5! = 120 words

N __ __ __ __ __ = 5! = 120 words

T __ __ __ __ __ = 5! = 120 words

Z E H __ __ __ = 3! = 6 words