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I have never played poker in my life and i have to solve this complicated problem.

How many five card poker hands contain at least $3$ jacks?

Here is what i know:

There are $52$ cards in a deck. A,2,3,4,5,6,7,8,9,10,J,Q,K;

$13$ cards and $4$ suits: diamonds, clubs, spades, and hearts.

5 poker hands:_ _ _ _ _

would be it $(52, 48)$ and then $(52, 26)$?

how would you computate it??

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  • $\begingroup$ Hint: Use the formula $\frac{n!}{k!(n-k)!}$, and then add i and ii $\endgroup$ – ammie Jun 15 '13 at 23:20
  • $\begingroup$ or n!/k!(n-k)! then you will be find it haha $\endgroup$ – ammie Jun 15 '13 at 23:34
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One needs to know nothing about poker. Indeed you need to know only that there are $52$ different cards in a standard deck, of which $4$ are Jacks.

We are asked how many $5$-card hands there are that contain at least $3$ Jacks.

We have at least $3$ Jacks if we have (i) exactly $3$ Jacks or (ii) exactly $4$ Jacks.

(ii) We count first the $4$-Jack hands. A hand has $5$ cards, so the number of $4$-Jack hands is the number of ways to pick the non-Jack card that will keep the four Jacks company. There are $48$ non-Jacks, so there are $48$ $4$-Jack hands. This looks a little nicer as $\binom{48}{1}$.

(i) We now count the $3$-Jack hands. Which Jacks are in the hand? They can be chosen in $\binom{4}{3}$ ways. For any choice of Jacks, the two non-Jacks can be chosen from the $48$ non-Jacks in the deck in $\binom{48}{2}$ ways. So the number of $3$-Jack hands is $\binom{4}{3}\binom{48}{2}$.

For the number of hands that have at least $3$ Jacks, add the counts obtained in (ii) and (i).

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