Q. 134.5( 2 Votes )
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?
Answer :
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.
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