Question 1173674
If there are 52 cards and 3 aces in a hand that means that there are 49 cards to choose 2 from for the rest of the 5 card hand. 

We can use the choose notation for this. 
The equation would be 49C2 which is the same as (49!)/(47!)/(2!). The way you get this the number of objects to choose from factorial or in this case 49! divided by the number of choices factorial or in this case 2! divided by the number of objects minus the number of choices factorial or in this case (49-2)! or 47!. Factorial is essentially the number times every number before it, so 4! would be 4*3*2*1 or 24. So (49!)/(47!)/(2!) is 1176 when we have 3 aces, but if we have 4 aces we only have 1 card to replace from the 5 card hand from 52-4=48 remaining cards. So only 48 options. In total there is 1176+48=1224 different hands of 5 cards containing at least 3 aces. 

The reason we used the factorial function for this type of problem is because when you input one card of the first 49, then you only have 48 left, then 47 and so on That's why we use factorial.