COMBINATION PROBLEMS WITH SOLUTIONS

Problem 1 :

Find the number of ways in which 4 letters can be selected from the word ACCOUNTANT.

Solution :

Problem 2 :

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?

Solution :

Number of white balls  =  2

Number of black balls  =  3

Number of red balls  =  4

Number of non black balls  =  2 + 4  =  6

Number of ways 

  =  (³C₁ ⋅ ⁶C₂) + (³C₂ ⋅ ⁶C₁)  +  (³C₃ ⋅ ⁶C₀)

  =  (3 ⋅ 15) + (3 ⋅ 6) + (1 ⋅ 1)

=  45 + 18 + 1

=  64

Problem 3 :

Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION?

Solution :

There are 11 letters not all different.

They are AA, II, NN, E, X, M, T, O.

The following combinations are possible:

Case 1 :

Number of ways selecting 2 alike, 2 alike

  =  ³C₂   =  3 ways

Case 2 :

Number of ways selecting 2 alike,2 different

  =  ³C₁ ⋅  ⁷C₂ ==> 3 x 21  ==>  63 ways.

Case 3 :

Number of ways selecting all 4 different =  ⁸C₄ 

= 70 ways.

Total number of combinations = 3 + 63 + 70 = 136 ways.

Total number of permutations (1) to (3)

  =  3 ⋅ (4!/2!2!) + 63 ⋅ (4!/2!) + 70 ⋅ 4!

=  18 + 756 + 1680

= 2454

Problem 4 :

How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?

Solution :

From the given question, we come to know that any three points are not collinear.

By selecting any three points out of 15 points, we draw a triangle.

Number of ways to draw a triangle  =  ¹⁵C₃

  =  (15 ⋅ 14 ⋅ 13) / (3 ⋅ 2 ⋅ 1) 

  =  455

Problem 5 :

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

Problem 6 :

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 number of ways is

= 126 + 84 + 36 + 9 + 1

= 256

Problem 7 :

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)

Number 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

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