SOLVED PROBLEMS ON COMBINATIONS

Solved Problems on Combinations :

In this section, we will learn, how to solve problems on combinations. 

Solved Problems on Combinations

Problem 1 :

There are 5 teachers and 20 students. Out of them a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees

(i) a particular teacher is included?

(ii) a particular student is excluded?

Solution :

Number of ways in which the commitee can be formed  =  

5C2   20C3

(i) a particular teacher is included?

When a particular teacher always included, it is enough to select the remaining 1 teacher out of 4 teachers.

Total number of ways  =  =  4C1   20C3.

(ii) a particular student is excluded?

When a particular student always excluded, it is enough to select the remaining 2 students out of 19 students.

Total number of ways  =  =  5C2   19C3.

Problem 2 :

In an examination a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways a student can answer the questions?

Solution :

Total number of questions  =  9

Number of questions must be answered  =  2

Number of questions to be answered  =  5

Number of ways to answer the questions  =  9 - 2  =   7C3

  =  35

Problem 3 :

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.

Solution :

Total number of cards in a deck  =  52

Number of ace cards  =  4

Number of cards to be selected  =  5

Here we must select 3 ace cards out of 4 and 2 other cards out of 48.

Number of ways  =  4C  48C2

  =  4 ⋅ [(48 ⋅ 47)/2]

  =  4 (1128)  =  4512

Hence the total number of ways is 4512.

Problem 4 :

Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority in the committee.

Solution :

7 Indians

5 Americans

Committee to be formed

5

4

3

0

1

2

5

5

5

Number of ways  =  (7C  5C0) + (7C  5C1) + (7C  5C2)

  =  (21 ⋅ 1) +  (35 ⋅ 5)  + (35 ⋅ 10)

  =   21 + 175 + 350 

  =  546

Hence the required number of ways is 546.

After having gone through the stuff given above, we hope that the students would have understood, how to solve problems on combinations

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Multiplicative Inverse Worksheet

    Jan 19, 22 09:34 AM

    Multiplicative Inverse Worksheet

    Read More

  2. Multiplicative Inverse

    Jan 19, 22 09:25 AM

    Multiplicative Inverse - Concept - Examples

    Read More

  3. Graphing Linear Functions Worksheet

    Jan 19, 22 08:24 AM

    Graphing Linear Functions Worksheet

    Read More