# Forming a committee but there must be more or equal amount of female to males

Suppose that a department contains 10 men and 15 women. How many ways are there to form a committee with six members if it must have at least as many women as there are men in it?

So my current approach on this question is to make 4 different cases where there are at least as many women as there are men. So I have:

Case 1: 6 Women, 0 Men

Case 2: 5 Women, 1 Men

Case 3: 4 Women, 2 Men

Case 4: 3 Women, 3 Men

For each case I have calculated the combinations as follows:

Case 1: $\binom{15}{6}$

Case 2: $\binom{15}{5} * \binom{10}{1}$

Case 3: $\binom{15}{4} * \binom{10}{2}$

Case 4: $\binom{15}{3} * \binom{10}{3}$

I'm not sure what I should do at this point or even if what I have done is correct.

• Everything you've done looks fine, just add up the cases and you're done. – user2566092 Mar 9 '15 at 23:50

You may want to evaluate your 'choose' functions (binomial coefficients) using the following formula: $${n \choose k} = { n! \over k! (n-k)! }$$