I have to count with 3 extractions without repetition from a deck of 52 cards (flop in texas holdem),
How many possible combinations of 3 cards that have an Ace?
How many possible combination of 3 card with one Ace, one card to 2-5(with different seeds) , one card to 6 to 9 (with different seeds).
I have try to resolve :
We have 3 possible position and in one we have ace.
A _ _
I can write A in 4 ways possible, Now deck have 52-4(ace) = 48 cards.
I think that we have 48 and 47 possible choices for the other two positions. Then $4 \cdot 48 \cdot 47 = 9024$ possible combination with an ace.
Second quest. I try :
scenario is A [2-5] [6-9]
we have 4 Ace,
we have 2-5 with different seeds
So, $4\cdot 4 = 16$ choices and 6-9 with different seeds = $4\cdot 4 = 16$ choices
I think that we have $4\cdot 16\cdot 16$ possible choices.
But I don't know if, for example A 2 7 is equal to 2 A 7 and if I counted it twice.