We're drawing 3 cards from a deck of 52, and we have
[tex]\dbinom{52}3=\dfrac{52!}{3!(52-3)!}[/tex]
ways of drawing any 3-card hand.
Of the 4 total aces in the deck, we want to draw 2. The third card can be any 1 of the 48 remaining cards. We have
[tex]\dbinom42\cdot\dbinom{48}1=\dfrac{4!}{2!(4-2)!}\cdot\dfrac{48!}{1!(48-1)!}[/tex]
possible 3-card hands that contain any 2 aces.
The probability of drawing such a hand is then
[tex]\dfrac{\binom42\cdot\binom{48}1}{\binom{52}3}=\dfrac{72}{5525}[/tex]
