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I'm trying to figure out two methods in doing these sorts of combination problems that show up on the GMAT. Mind helping me out in understanding both methods?

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?

1) Why does this method work? 6C3 * 3C2 = 6!/(3! * 3!) * 3!/(1! * 2!) = 20 * 3 = 60

60 I believe is correct. But why do we multiply? What is going on?

2) 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. But how do I do it?

I'll start off.

9C5 = 9!/(5! * 4!) = 126

Jwan622
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1 Answers1

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We multiply because of the "and": you want 3 guards AND 2 forwards. This is sometimes called the rule of product.

The way you did it in (1) is the easiest way, IMO:

How many 3 guard combinations are there from 6 available guards? $C(6,3)={6!\over 3!3!}=20$.

How many 2 forward combinations are there from 3 forwards? $C(3,2)=3$.

Therefore, there are $C(6,3)\cdot C(3,2)=20\cdot 3=60$ possible teams.

JohnD
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