Solution
We need to remember that in a standard deck we have 4 aces and 48 other types of cards
For this case we can solve the problem with the follwoing operation:
[tex](4C2)\cdot(48C3)=\frac{4!}{2!2!}\cdot\frac{48!}{45!3!}=6\cdot17296=103776[/tex]So then we have 103776 ways to create the combination required
