Probability of having only 2 suits represented in a hand of 5 cards. I know the denominator will be (52 c 5) to represent the total number of 5 card hands with 52 cards. I'm having trouble with the numerator. I began with (4 c 2) for the total number of combinations of 2 suits of the 4. Then I multiplied this by 16 (the number of ways to arrange 2 suits in 5 slots, which would be 2 x 2 x 2 x 2 x 1). Am I on the right track so far? I know I'm going to have to multiply this by the probabilities of drawing those suits, like 13/52 and such, but I'm not sure how to do this correctly.
 A: My thinking...(with assumption both suits must be represented)
You start with your suggestion to use $4\choose2$ to get the two suits.  Then from two suits, there are $26\choose5$ ways to make a hand.  Then, subtract out the $13\choose5$ hands that are only the first suit and $13\choose5$ hands that are only the second suit.  Which equals David's summation method.
$${4\choose2}({26\choose5}-2{13\choose5})$$
A: Our Sample Space is $\binom{52}{5}$ now how do we create a hand of cards with only 2 suits in it? The first step would be to choose two suits, $\binom{4}{2}$ ways to do that. Now, we discard all the cards which are not of that suit (1 way to do that). We are now left with 26 cards, 13 of each suit. Now we can break it into cases.
Case I: We have 1 card of suit 1 and 4 cards of suit 2
$\binom{4}{2}*\binom{13}{1}*\binom{13}{4}$ ways to make that kind of hand
Case II: We have 2 card of suit 1 and 3 cards of suit 2
$\binom{4}{2}*\binom{13}{2}*\binom{13}{3}$ ways to make that kind of hand
Case III: We have 3 cards of suit 1 and 2 cards of suit 2
$\binom{4}{2}*\binom{13}{3}*\binom{13}{2}$ ways to make that kind of hand
Case IV: We have 4 cards of suit 1 and 1 card of suit 2
$\binom{4}{2}*\binom{13}{4}*\binom{13}{1}$ ways to make that kind of hand
If we add all those cases up, we have all possible hands which are represented by 2 suits exactly. Divide these by $\binom{52}{5}$ and you'll get your answer, OP.
A: I'm gonna propose a solution that I feel is simpler. There is a 1 in 4 chance that two cards are the same and a 3 in 4 chance that they are different. If they're different, then for each card that is added there's a 2 in 4 chance that it will be of the same suit as one of the first two cards ($0.75\cdot0.5\cdot0.5\cdot0.5=0.09375$). If the first two cards are the same, then you can just repeat the first step. This leads to several probabilities, and then the answer is just the union of those probabilities.
$$0.75\cdot0.5\cdot0.5\cdot0.5\cup0.25\cdot0.75\cdot0.5\cdot0.5\cup0.25\cdot0.25\cdot0.75\cdot0.5\cup0.25\cdot0.25\cdot0.25\cdot0.75\cup0.25\cdot0.25\cdot0.25\cdot0.25=0.1796875=\frac{23}{128}$$
(If you don't want to include the case where all cards are a single suit, just cut out the last part)
