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Here's the question:

A 5-card hand is selected from a standard deck of playing cares. (A standard deck has 13 cards from each of 4 suits{clubs, diamonds, hearts, and spades. The 13 cards have face value 2 through 10, jack, queen, king, or ace. Each face value is \kind" of card. The jack, queen, and king are \face cards.") How many hands contain exactly one pair?

And here's the provided answer:

We can view this as a sequence of events. First, select the face value for the pair. There are 13 choices for this. Next, select which two cards of that face value will make up the pair. Thus, we want to select 2 cards from the 4 of that face value at this step. There are C(4; 2) = 6 ways to do this. Third, we need to select the other three cards. These other three cards cannot contain any pairs. There are 48 ways to select the rest of these cards, since we have to exclude the cards with the face value of the pair. The next card has 44 options, since we cannot select any of the two face values used thus far. The third of these three cards has 40 options, since we cannot use any of the 3 face values already used. However, we are over counting by all possible ways to arrange these three cards, so we must divide out by the number of arrangements, namely 3! = 6. The final answer is: (13*6*48*44*40)/6 = 1098240

I understand all of the steps except one: how are we over counting? P(3,3) = 6 which implies that the ordering of the cards in the hand matters, which makes no sense to me because it's still the same hand regardless of what order it's in. So how are we over counting, where is 3!=6 coming from if not P(3,3).

Thanks!

OSFTW

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Order does not matter, as you say. However, we counted the last three cards as if order does matter. We correct for this by dividing by $3!$.

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  • $\begingroup$ So, by selecting one card, then the next card, then the next card (as opposed to say all 3 at once which isn't possible since we need to make sure no two are the same kind), we counted as if order mattered?? I think that makes sense, so we'll have to account for that by dividing by 3!. Could you go into a bit more explanation about how we count the 3 cards as if order mattered? Thanks! $\endgroup$ – OpenSrcFTW Nov 15 '13 at 0:15
  • $\begingroup$ We counted Kh, Qs, Jd and we also counted Jd, Kh, Qs and also four other permutations. But these are really all the same. Similarly for every other triple; we counted each one six times. $\endgroup$ – apt1002 Nov 15 '13 at 0:27
  • $\begingroup$ Kh, Qs, Jd?? What do you mean by those? $\endgroup$ – OpenSrcFTW Nov 15 '13 at 2:04
  • $\begingroup$ King of hearts, Queen of spades, Jack of diamonds. $\endgroup$ – apt1002 Nov 15 '13 at 2:20
  • $\begingroup$ Oh ok, thanks! Answer selected. $\endgroup$ – OpenSrcFTW Nov 15 '13 at 4:01

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