# A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions?

Given that we need to answer 7 questions from 2 groups which consist of 6 questions each.

It is also told the candidate is permitted to answer the utmost 5 questions from any group.

The cases for answering the 7 questions:

i. 5 questions from group 1 and 2 from group 2

ii. 4 questions from group 1 and 3 from group 2

iii. 3 questions from group 1 and 4 from group 2

iv. 4 questions from group 1 and 5 from group 2

Let us assume the total no. of ways of answering 7 questions be N.

N = no. of ways of answering 7 questions from both groups

N = (No. of ways of answering 5 questions from group 1 and 2 from group 2) + (No. of ways of answering 4 questions from group 1 and 3 from group 2) + (No. of ways of answering 3 questions from group 1 and 4 from group 2) + (No. of ways of answering 2 questions from group 1 and 5 from group 2)

N = (6C5 × 6C2) + (6C4 × 6C3) + (6C3 × 6C4) + (6C2 × 6C5)

We know that ,

And also n! = (n)(n – 1)......2.1

N = (6 × 15) + (15 × 20) + (20 × 15) + (15 × 6)

N = 90 + 300 + 300 + 90

N = 780

The total no. of ways of choosing 7 questions is 780 ways.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Permutations & Combinations | Analyze your learningFREE Class
Challenging Quiz on P&C | Test Yourself55 mins
Lecture on Combinations49 mins
Understand Permutations like never before60 mins
Check Your progress Part 2| Interactive Quiz: Permutation & Combination54 mins
Interactive Quiz on Division and distribution of objects17 mins
Combinations - BOX & GAP Method38 mins
Interactive Quiz on Combinations50 mins
Lets see how good you can manage numbers | P&C Quiz Challenge41 mins