# exercise on combinations (poker deck)

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.

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 $$\binom{4}{1}\binom{16}{1}\binom{16}{1}$$