Probability/Combinatorics Problem - Old Maid Cards 
A special deck of Old Maid cards consist of 25 pairs and a single old maid card.  All 51 cards evenly between you and two other players – 17 cards for each player.
(a)  how many different hands can be dealt to you?
(b)  what is the probability that your hand has exactly 2 pair (and 13 single cards)?

(a) This is easy:  $\binom{51}{17}$
(b)  This one I'm having trouble.  I thought about doing something like this:
$\cfrac{\binom{25}{2}*\binom{47}{13}}{\binom{51}{17}}$
25-C-2 = Choose 2 of 25 pairs
47-C-13 = There's 46 remaining cards (or 23 pairs) but you add 1 because of old maid card
51-C-17 = Total possibilities.
I know this answer is wrong because its greater than 1.  The solution is 0.30282.
Any help is appreciated.  Thank you.
 A: Let’s split the calculation into two cases, hands with the Old Maid and hands without.
A hand with the Old Maid that has exactly two pairs contains the Old Maid, two pairs, and $12$ singletons. Such a hand therefore has cards from $12+2=14$ of the $25$ denominations. There are $\binom{25}2$ ways to choose the denominations of the two pairs, and there are then $\binom{23}{12}$ ways to choose the denominations of the $12$ singletons. For each singleton there are $2$ ways to choose which member of the pair we get. Thus, there are $2^{12}\binom{25}2\binom{23}{12}$ hands of this type.
Now see if you can modify that calculation to get the number of hands of the desired type that do not include the Old Maid.
A: I was able to answer this with Brian M. Scott's post (thanks Brian)!  This problem must first be decomposed into 2 cases:
(1)  Hand with the old maid card:  Pair1, Pair2, Old Maid, 12 denominations.
(2)  Hand without the old maid card:  Pair1, Pair2, 13 denominations.
Case 1:
Number of denominations= 12 singles + 2 pairs = 14 denominations.  The two pairs can be chosen $\binom{25}{2}$ ways.  The remaining 12 single denominations can be chosen $\binom{23}{12}$ ways by the partition principle.  However, for each single denomination, you have 2 choices so you multiply by $2^{12}$.  
$\binom{25}{2} \binom{23}{12} 2^{12}$
Case 2:
Number of denominations= 13 singles + 2 pairs = 16 denominations.
By the same logic, the number of outcomes is= $\binom{25}{2}\binom{23}{13} 2^{13}$
Solution:
$Pr=\cfrac{\binom{25}{2} \binom{23}{12} 2^{12} + \binom{25}{2}\binom{23}{13} 2^{13}}{\binom{51}{17}}=0.3028278$
