Boring Gin Question

[LetYouDown]

Stole this from a PDF I was pointed to. Figure with the lack of content and gin posts...I'd post something to kill about 1.5 minutes of everyone's boring work day.

A gin hand consists of 10 cards from a deck of 52 cards. Find the probability that a gin hand has
(a) all 10 cards of the same suit.
(b) exactly 4 cards in one suit and 3 in two other suits.
(c) a 4, 3, 2, 1, distribution of suits.

[TomCollins]

a) C(4,1) * C(13,10) / C(52,10) = 7.23134165 × 10-08
b) C(4,1) * C(13,4) * C(3,1) * C(13,3) * C(2,1) * C(13,3) / C(52,10) = 0.0887242232
c) C(4,1) * C(13,4) * C(3,1) * C(13,3) * C(2,1) * C(13,2) * C(1,1) * C(13,1) / C(52,10) = 0.3145677

edit: thx google calculator
edit: I may have screwed something up, probably did. Someone find my dumb mistake.

[TomCollins]

Found my mistake:

b) C(4,1) * C(13,4) * C(3,2) * C(13,3) * C(13,3) / C(52,10)
c) C(4,1) * C(13,4) * C(3,1) * C(13,3) * C(2,1) * C(13,2) * C(1,1) * C(13,1) / C(52,10)