Question

A candidate is required to answer 6 out of 12 questions which are divided into two parts A and B, each containing 6 questions and he/she is not permitted to attempt more than 4 questions from any part. In how many different ways can he/she make up his/her choice of 6 questions ?

• A. 850
• B. 800
• C. 750
• D. 700

Answer 1

The Correct Option is A

$A\,\,\,B$
$4\,\,\,2\,\,\,\to\,\,\,^{6}C_{4} \times \,^{6}C_{2}$
$2\,\,\,4\,\,\,\to\,\,\,^{6}C_{2} \times \,^{6}C_{4}$
$3\,\,\,3\,\,\,\to\,\,\,^{6}C_{3} \times \, ^{6}C_{3}$
Total $2\times \, ^{6}C_{2}\times \, ^{6}C_{4}\times \, ^{6}C_{3}\times \, ^{6}C_{3} = 450 + 400 = 850$

Answer 2

We have to find the total number of ways he/she makes up his/her choice of 6 questions with the condition that he/she can choose at most 4 questions from either part A or B. So, here we have three cases:

Case-1:

If he/she chooses to answer 4 out of the required 6 questions in part A and rest 

6−4=2 questions in part B. Then he/she can select 4 questions from available 6 questions in part A in ⁶C₄ ways and then he/she could also select 2 questions from available 6 questions in part B in ⁶C₂. Therefore, we use the rule of product and find the number of ways he/she can choose to answer 6 question in the case-1 as follows:

N₁=⁶C₄×⁶C₂

Case-2:

If he/she chooses to answer 3 out the required 3 questions in part A and rest 

6−3=3 questions in part B. Then he/she can select 3 questions from available 6 questions in part A in ⁶C₃ ways and then he/she could also select 3 questions from available 6 questions in part B in ⁶C₃. Therefore, we use the rule of product and find the number ways he/she can choose to answer 6 questions in case-2 as follows:

N₂=⁶C₃×⁶C₃

Case-3:

If he/she chooses to answer 2 out the required 6 questions in part A and rest 

6−4=2 questions in part B. Then he/she can select 2 questions from available 6 questions in part A in ⁶C₂ ways and then he/she can also select 4 questions from available 6 questions in part B in ⁶C₄. Therefore, we use rule of product and find the number ways he/she can choose to answer 6 questions in case-3 as follows:

N₃=⁶C₂×⁶C₄

Hence, we can see that he/she can choose 6 questions following either case-1, case-2 or case-3. So now, we use rule of sum and find the total number of ways he/she choose 6 questions as follows:

N₁+N₂+N₃=⁶C₂×⁶C₄+⁶C₃×⁶C₃+⁶C₄×⁶C₂

⇒ N₁+N₂+N₃=\(\frac{6!}{4!2!}\times\frac{6!}{2!4!}+\frac{6!}{3!3!}\times\frac{6!}{3!3!}+\frac{6!}{2!4!}\times\frac{6!}{4!2!}\)

⇒ N₁+N₂+N₃=15×15+20×20+15×15

⇒ N₁+N₂+N₃ = 225+400+225 = 850

Therefore, the correct answer is “Option A” i.e., 850.