I feel the answer should be (7!/(5!2!))^3 or 7C2*7C2*7C2, not (7!/(5!2!))3

selectin 2 from 7 is 7C2
NOW 7C2==7!/5!*2!
similalrly 3 level Q means 7C2*7C2*7C2
HENCE a

Can someone please explain in more detail why a is correct?

So there are 21 questions total which are grouped into three groups (hard, intermediate, & easy) of seven questions each. You are then told to find the number of ways to pick 2 questions from EACH group (each group has seven questions total).

Because order does NOT matter, this problem is a combination, so you will use "C" NOT "P". So,

Group 1 (Hard questions): 7 questions total. You are choosing 2 --> 7C2 --> 7!/(5!2!)
Group 2 (Intermediate questions): 7 questions total. You are choosing 2 --> 7C2 --> 7!/(5!2!)
Group 3 (Easy questions): 7 questions total. You are choosing 2 --> 7C2 --> 7!/(5!2!)

The next step is then to multiply NOT add. The answer can be written out in simple english as "How many ways can I pick 2 questions from group 1 (hard) AND pick 2 questions from group 2 (intermediate) AND I pick 2 questions from group 3 (easy)?" Multiplication essentially stands for the AND part. Hope that makes sense.

So, the answer is 7C2 * 7C2 *7C2 which is the same as (7C2)^3 which is the same as (7!/(5!2!))^3

As the answer is 7C2 the answer should be B as it states that 7!/5!^3 i.e

7*6*5*4*3*2*1/5*4*3*2*1


so that leave us with 7*6 and therefore B


Please tell me what mistake I'm doing


Thank You

mehulv,

your mistake is that you think 7C2 is 7P2.

Any combination such that rCn = r!/(r-n)!n!. Any permutation such that rPn = r!/(r-n)!

So, 7C2 is correct. The problem is that you wrote it out as a permutation. Make sure to memorize the permutation and combination statements again.

To reiterate, 7C2 = 7!/5!2! = (7*6*5*4*3*2*1)/(5*4*3*2*1*)(2*1)