Finite Math
posted by Zoe .
A "hand" consists of a set of 5 cards selected from an ordinary deck of 52 cards:
(1) How many different hands contain exactly 4 diamonds? = 27885
(2) How many different hands contain exactly 4 cards of the same suit (any of the four suits)?
(3) How many different hands contain cards of exactly one suit?
Can someone help me figure out the last two questions?

so you want 4 out of the 13 diamonds > C(13,4)
then one more card from the 39 nondiamonds > C(39,1)
so 715x39 = 27885
2. wouldn't the answer to 1. which was for diamonds be the same for the clubs, the same for the hearts, and the same for the spades ???
so 4x27885
3. I assume the suit would be specified.
If not, then for each specific suit is would be C(13,5) = 1287 , that is, all 5 cards come from the same 13 cards. 
Thanks Reiny!!!
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Can someone please explain this to be, so that I can understand this problem?