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 = (^{6}C_{5} × ^{6}C_{2}) + (^{6}C_{4} × ^{6}C_{3}) + (^{6}C_{3} × ^{6}C_{4}) + (^{6}C_{2} × ^{6}C_{5})

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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