TITLE: Number of hand of cards with exactly 3 aces (Unsure if answer is correct)
QUESTION [6 upvotes]: Problem
A pokerhand is 5 card subset, which is picked from 52 cards in total. Four of all cards are aces. Now how many there are such pokerhands that contain exactly 3 aces and 2 cards that can be anything. Also what is probability for obtaining exactly  
Attempt to solve
Now we can pick 3 aces from total of 4 aces and pick any two cards that are not aces.
$$ (\text{number of possible aces})(\text{number of possible not aces}) $$
$$ {{4}\choose{3}} {{48}\choose{2}}=54155$$
Now the probability would be simply:
$$ \frac{\text{number of hands with aces}}{\text{number of all hands}} $$
$$ \frac{{{4}\choose{3}} {{48}\choose{2}}}{{{52}\choose{5}}}\approx 1.736079047\cdot 10^{-4}$$
We get very small probability of $\approx 0.017\%$

Combinatorics isn't strong point of mine so if someone could point out possible flaws that there are with my approach that would be highly appreciated. Also if the approach seems correct please comment that this seems correct.

REPLY [2 votes]: This is the correct approach to the problem. A nitpick though, it should be:
$$\text{total ways to choose such pokerhand}=\text{number of ways to choose 3 aces}\cdot\text{number of ways to choose 2 other cards}$$
I assume you had the same intention in mind but couldn't express it in words. 
Apart from it, rest all is correct! You also posted a nicely formatted question. Good job!