Definitely not sure about this one but wouldn't it be 60?

different ways to get guards 6!/(3! * 3!) = 20

different ways to get forwards 3!/2! = 3

20 * 3 = 60

Yes you are right ! its 6C3 * 3C2 = 60. But is it really 700-800 level problem?

osirus0830 wrote:Definitely not sure about this one but wouldn't it be 60?

different ways to get guards 6!/(3! * 3!) = 20

different ways to get forwards 3!/2! = 3

20 * 3 = 60

sumanr84 wrote:9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
23
30
42
60
126

MGMAT ( 700-800)

3 Guards can be selected from 6 in 6C3 = 20 ways
2 Forwards can be selected from 3 in 3C2 = 3 ways

Total no of ways 20*3 = 60

GMAT 700-800 level? are you kidding me?

6C3*3C2

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.

sumanr84 wrote: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.

Such a useful explanation after test.