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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.

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How many possible combinations of $3$ cards include an ace?

Method 1: We make a direct count.

There are $4$ aces and $52 - 4 = 48$ non-aces in a standard deck. A selection that includes an ace must have one, two, or three aces. If it has exactly $k$ aces, then it must include $48 - k$ of the other $48$ cards in the deck. Hence, the number of three-card hands that include an ace is $$\binom{4}{1}\binom{48}{2} + \binom{4}{2}\binom{48}{1} + \binom{4}{3}\binom{48}{0}$$ where $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ is the number of ways of selecting a subset with $k$ elements from a set with $n$ elements.

Method 2: We count indirectly.

There are $\binom{52}{3}$ ways to select three of the $52$ cards in the deck. Of these, $\binom{48}{3}$ contain no aces. Hence, the number of selections that include an ace is $$\binom{52}{3} - \binom{48}{3}$$

How many combinations of three cards include an ace, a card numbered from $2$ to $5$, and a card numbered from $6$ to $9$?

It is not clear what you mean by different seeds. However, if this is the question that you intended to ask, you have solved it correctly. Since there are four aces, $16$ cards numbered from $2$ to $5$, and $16$ cards numbered from $6$ to $9$, the number of ways of selecting one of each type is $$\binom{4}{1}\binom{16}{1}\binom{16}{1}$$

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