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I am a little bit confused about this problem. It said that three cards are dealt from a standard deck of cards. In how many ways can one get at least one king?

Now If I use the complement, then I want the number of ways one can deal 3 cards minus the number of ways of not getting a king. Well, 52*51*50 = 132600 is the number of ways of dealing 3 cards. The number of ways not dealing king is 48*47*46 = 103776. So I'd 132600 -103776 = 28824 ways of getting at least one king. Is this reasoning right?

On the other hand if I do $\binom{4}{1}\binom{48}{2} + \binom{4}{2}\binom{48}{1} + \binom{4}{3}\binom{48}{0}$, I get a different answer. Am I doing something wrong?

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Your first count takes into account the order in which the cards are dealt; so dealing ace-clubs/2-hearts/3-diamonds counts as a different "deal" from 2-hearts/3-diamonds/ace-clubs. Your second method does not take into account the order, just what the three cards in question are. So that's why you are getting different answers. –  Arturo Magidin May 31 '12 at 20:24
    
Okay. So which one is the correct solution in this case? –  count May 31 '12 at 20:26
    
@count: Depends on the problem. Does the problem consider the order of the cards, or just the final deals? If the former, then the first solution; if the latter, the second solution. –  Arturo Magidin May 31 '12 at 20:27
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The simplest interpretation is that you are looking at the completed deal, not at the order in which the cards are dealt, so I would say that the correct count is the second one. –  Arturo Magidin May 31 '12 at 20:37
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The key phrase is "in how many ways." Given that "dealing" three cards usually involves the sequential giving of cards, the phrase "in how many ways" to me indicates that order matters. But the question is poorly phrased, making the author's intent far from obvious. –  Thomas Andrews May 31 '12 at 20:46
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1 Answer 1

The second result ist correct. The first one is not, since you're counting some configurations more than once.

I'd say the number of ways you can deal 3 cards is $\frac{52*51*50}{3*2*1}=22100$, the number of ways you can deal 3 cards with no king is $\frac{48*47*46}{3*2*1}=17296$. So the difference is $4804$, which agrees with the second result

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