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I am stomped on the following question

How many different ways are there to draw $6$ cards from a standard deck of cards and obtain $4$ kings and $2$ jacks? (The Answer is $6$)

I believe I am starting the question all wrong since I am doing this for

How many different ways are there to draw $6$ cards from a standard deck of cards

No. of ways = $\frac{52!}{46!6!} $

Any suggestions on how I would solve this problem ?

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This is about combinations, not permutations. –  Michael Hardy Aug 6 '12 at 19:32
    
@MichaelHardy Just fixed the tag. Thanks. –  MistyD Aug 6 '12 at 19:33

2 Answers 2

up vote 2 down vote accepted

I believe the total number of hands is completely irrelevant for answering this question. All you need to know is that you need 6 cards. 4 cards need to be kings. 4 cards need to be jacks. You only have 4 kings, and you only have 4 jacks. The number of ways you can draw 4 kings out of a set of 4 kings is 4C4 = 1. The number of ways you can draw 2 jacks from a set of 4 is 4C2 = 6.

This makes your answer 4C4 * 4C2 = 1 * 6 = 6

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I think you got the formulas mixed up –  sidht Aug 6 '12 at 19:12
    
@jak sorry u are right... –  mathguy Aug 6 '12 at 19:13
    
@Sidd regarding your previous statement in which you stated that $\frac{52!}{46!6!}$ gives us a combination of 6 cards from 52 cards and that irrelevant here. Arent we exactly told to do that How many different ways are there to draw 6 cards from a standard deck of cards –  MistyD Aug 6 '12 at 19:16
    
@MistyD Oh sorry. I thought you were saying that you needed the answer to that question to answer the first question. If it is a separate question, then you are doing it perfectly fine. The equation you wrote is the number of 6-card hands you can have from a 52-deck of cards. This is 52C6. –  mathguy Aug 6 '12 at 19:19
1  
That's because all we need to compute the answer is the kings and the jacks. The aces, 2-10, and queens aren't involved here at all. The formula you used to solve the second equation though (52C6) is used in the first question however (the answer is 4C4 * 4C2). If you were finding the probability of drawing 6 cards with 4 kings and 2 jacks, then you would use 52C6 in the denominator to replicate (number of needed events/number of possible events) for the probability –  mathguy Aug 6 '12 at 19:35

There is only $1$ way to select the kings. There are $\binom{4}{2}$ ways to select $2$ Jacks from the $4$ available.

Or else, in desperation, one can make an explicit list and count: (J$\spadesuit$ J$\heartsuit$); (J$\spadesuit$, J$\diamondsuit$); (J$\spadesuit$, J$\clubsuit$; (J$\heartsuit$, $J\diamondsuit$); (J$\heartsuit$, J$\clubsuit$); (J$\diamondsuit$, J$\clubsuit$).

Remark: As pointed out by jak, in order to prepare for more complicated problems, it may be better to say that there are $\binom{4}{4}$ ways to select the Kings, and for each of these ways there are $\binom{4}{2}$ to choose the Jacks, for a total of $\binom{4}{4}\binom{4}{2}$.

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Or we can say there is $\binom{4}{4}$ ways to choose the kings (since we are taking EXACTLY four kings) –  sidht Aug 6 '12 at 19:09

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