a group of 8 women and ten men must select a five person committee. how many committees are possible if more than half of the members are men?

Question

anonymous · Answer

(3M and 2W) + (4M and 1W) + (5M and 0W)

use combination formula

anonymous · Answer

hmm okay Im still not getting it

anonymous · Answer

maybe I did something wrong...

mathmale · Answer

sourwing:  I agree:  acceptable choices could be 3 men and 2 women, 4 men and 1 woman, or 5 men, no women.  What's your justification for writing (3M and 2W) + (4M and 1W) + (5M and 0W) as a sum?  

Would it not be correct to calculate how many combinations of 3 men and 2 women there are, and then how many of 4 and 1, and how many of 5 and 0, and then adding together the 3 resulting numbers of combinations?

anonymous · Answer

that's what I meant when I said use combination formula.
for 3M and 2W, it's (10C3) (8C2)

similar calculator for the rest

anonymous · Answer

maybe i'll just give the answer
(10C3) (8C2) + (10C4) (8C1) + (10C5) (8C0)

anonymous · Answer

thats the answer I dont know how to figure that out

anonymous · Answer

I just need it real quick! lol