Deal 5 cards to Player1 and 5 cards to Player2 from a deck of 52 cards What is the probability that :
a) Player1 has 2 aces
b) Player2 has 2 kings
c) Player1 has 2 aces and Player2 has 2 kings  
For a) I did :
$$\frac{2! \cdot {{50}\choose{3,3}} }{ {52}\choose{5,5} }$$
But I think that's wrong.
And I know that the result of a) will equal b).
And I have no clue for c)
 A: Let's say we deal to Player 1 first. Out of the 4 aces in the deck he gets exactly 2. There are $\binom{4}{2}$ many ways to choose the aces. Out of the 48 non-ace cards he gets 3. There are $\binom{48}{3}$ ways to choose those. Out of the remaining 47 cards (2 aces and 45 non-aces) Player 2 gets 5 cards. Thus the probability that Player 1 gets exactly 2 aces is:
$$
\frac{\binom{4}{2}\binom{48}{3}\binom{47}{5}}{\binom{52}{5}\binom{47}{5}} = \frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}.
$$
Note that Player 2 doesn't really matter for this.
As you already said, b) is just the same. See if you can do c) now.
Edit: Answer to c):
There are $3$ cases: Player $1$ has no king, one king or two kings. Let $i \in \{0,1,2\}$ be the number of kings in Player $1$'s hand. There are $\binom{4}{2}$ many ways to choose the $2$ aces for Player $1$ and $\binom{4}{i}$ many ways to choose the $i$ kings that Player $1$ has, the remaining $3-i$ cards of Player $1$ come from the $44$ cards that are neither an ace nor a king. Then for Player $2$ there are $\binom{4-i}{2}$ ways to choose the kings and finally we need to choose the remaining $3$ cards from the cards that are not in Player $1$'s hand and not kings (There are $52-5-4+i = 43+i$ such cards). Hence the probability is 
$$
\frac{1}{\binom{52}{5}\binom{47}{5}}\sum_{i=0}^2\binom{4}{2}\binom{4}{i}\binom{44}{3-i}\binom{4-i}{2}\binom{43+i}{3}.
$$
