There are 12 women and 16 men in a room. How many different committees of size 5 can be selected if the committee must have exactly 2 women and 3 men?
My answer makes no sense to me. I got 110, 880 but that seems far fetched. I did C(12,2)C(16,3)

Question

There are 12 women and 16 men in a room. How many different committees of size 5 can be selected if the committee must have exactly 2 women and 3 men?
My answer makes no sense to me. I got 110, 880 but that seems far fetched. I did C(12,2)C(16,3)

kropot72 · Answer

There are 12C2 ways of selecting the two women and 16C3 ways of selecting the two men. Total number of ways of selecting committee is given by 12C2 + 16C3

anonymous · Answer

So then my answer is actually 560?

anonymous · Answer

Excuse me- I mean 626?
12!/(12-2)!*2!= 66
16!/(16-3)!*3!=560
66+560= 626?

kropot72 · Answer

Yes, 626 is correct.

anonymous · Answer

Awesomesauce! Thank you!

kropot72 · Answer

You're welcome :)

kropot72 · Answer

Very sorry, my bad. The total number of ways of forming the committee of 5 is
$$12C2\times 16C3 = 66\times 560=36960$$