52 cards, 5 picked, 3 of same suit 5 cards are drawn from a normal deck of cards (52). What is the probability that 3, and only 3, of the cards are of the same suit?
I'm wondering if my reasoning is sound:
C(13,1) * C(4,3) * C(12,2) * C(4,1)^2
-------------------------------------
              C(52,2)

 A: This solution is assuming that there must be exactly $3$ cards with the same suit.
First, pick a suit, there are $4$ ways of doing this. Next, of the $52/4 = 13$ cards in that suit, pick $3$ of them. There are ${13 \choose 3}$ ways of doing this. Next, of the $52-13 = 39$ cards whose suit is not the one that is picked, pick $2$ of those cards. There are ${39 \choose 2}$ ways of doing this. Now you have a 5-card hand with the specified conditions. Thus, there is a total of $4 \cdot {13 \choose 3} \cdot {39 \choose 2}$ ways of picking a 5-card hand with the specified conditions. The total number of ways of picking a 5-card hand is ${52 \choose 5}$ thus the probability is given by,
$$\frac{4 \cdot {13 \choose 3} \cdot {39 \choose 2}}{{52 \choose 5}}.$$
A: 3 solutions.
Solution1: Exclude 1 suit from 52
Solution2: Consider cases for pair or not a pair for the 2 cards left
Solution3: Consider cases for cards of same suit or not of the same suit for the 2 cards left
Solution2:
4 ways for choose 3 cards from 13. (same suit)
for remaining 2 cards, we have 2 cases to consider:
1. can either have a pair. ( 3 choose 2) from 3 suits left.
2. not have a pair. (3 choose 1) 3 suits left.  
$$\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot \left( {13 \choose 2} \cdot {3 \choose 1}^2 + {13 \choose 1 } \cdot {3 \choose 2}\right)  }{{52 \choose 5}}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (702+39)}{52 \choose 5}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (741)}{52 \choose 5}$$
this would give the same solution as $${39 \choose 2} = 741$$
(exclude 1 suit from 52 cards) 
Solution3:
$$\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot \left( {13 \choose 2} \cdot {3 \choose 1} + {13 \choose 1 }^2 \cdot {3 \choose 2}\right)  }{{52 \choose 5}}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (234+507)}{52 \choose 5}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (741)}{52 \choose 5}$$
this would give the same solution as $${39 \choose 2} = 741$$
(exclude 1 suit from 52 cards) 
A: I think the answer you were looking for would make sure that the remaining two cards would also be of a different suit. This would just be a slight modification of benguin's answer, as he assumes that the remaining two cards could be of the same suit. From the cards remaining, there are  $3 \cdot {13 \choose 2}$ ways to choose two cards from the same suit, because there are 3 suits and 13 cards in each suit. Hence, the overall probability becomes 
$$\frac{4 \cdot {13 \choose 3} \cdot \left({39 \choose 2} - 3 \cdot {13 \choose 2}\right)}{{52 \choose 5}}$$
Hope that helps.
