# Math Help - Perm/Combinations problem

1. ## Perm/Combinations problem

A class of 40 students must form groups of five for a class project. There are 24 women and 16 men in the class and each group must include 3 women and 2 men. In how many ways can the groups be made?

The answer is 242,880 but I haven't clue on how to set this up

2. Hi, sarahh

С - combination Combination - Wikipedia, the free encyclopedia

C[n,k] = n!/(k!(n-k)!)

C[16,2] * C[24,3] = 242880

3. Originally Posted by sarahh
A class of 40 students must form groups of five for a class project. There are 24 women and 16 men in the class and each group must include 3 women and 2 men. In how many ways can the groups be made?
The answer is 242,880 but I haven't clue on how to set this up
That happens to be the number of ways to form one group of 3 women and 2 men.
But that is not what the question states. It says groups.
Is that a mistake?

4. Originally Posted by Plato
That happens to be the number of ways to form one group of 3 women and 2 men.
But that is not what the question states. It says groups.
Is that a mistake?
If what Plato says is correct then the answer should be
$
\frac{24!}{{(3!)}^{8}(8!)}. \frac{16!}{{(2!)}^{8}(8!)}
$

5. Thanks guys--the first way that pankaj solved it must be right since that would be the solution that is given..however, my instructor must have been clumsy with the language. Many thanks again!