I have no idea why it was ranged as 700-800 in MGMAT CAT 5.
Here is how their solution goes,
This is a constrained combinatorics problem. One way to calculate the answer would be to compute the number of different teams of 5 that could be chosen out of 9 players, and then subtract out all of the teams that have too many guards or too many forwards. However, that approach will get messy.
The trick to doing this problem fairly easily is to recognize that we aren't choosing a team of 5 players (out of 9) at all: we're actually choosing a team of 3 guards (out of 6) and, SEPARATELY, a team of 2 forwards (out of 3). This is called a MULTI-STAGE combinatorics problem, where we can separate the original "pool" of players into sub groups and choose from the sub groups in different stages. First choose 3 guards from a pool of 6 guards. Next, choose 3 forwards from a pool of 3 forwards.
There are 6 guards competing to be on the team, and the team will consist of 3 guards. Using the anagram method, we get:
1 2 3 4 5 6
Y Y Y N N N
Therefore. the number of different combination is:
6!/(3!3!) = 20
There are 20 different combinations of guards that can be chosen for the basketball team.
Similarly, there are 3 forwards competing to be on the team, and the team will consist of 2 forwards. Using the anagram method, we get:
1 2 3
Y Y N
Therefore. the number of different combinations is:
3!/(2!1!) = 3
There are 3 different combinations of forwards that can be chosen for the basketball team.
We now need to multiply these numbers together: we have 20 different options for our "guards" portion of the team, and each of these 20 options can be paired with any of our 3 different options for the "forwards" portion of the team. Therefore, there are 20 × 3 = 60 different possible teams of 3 guards and 2 forwards that can be chosen.
The correct answer is D.
I am on a break !!
