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Problem Poker dice is played by simultaneously rolling 5 dice. How many ways can we form "1 pair", "2 pairs"?

For one pair, I got the answer right away. First I consider there are 5 spots for 5 dices. Then I pick 2 places out of 5, which means there are 3 places left, so we have to choose 3 out of 3 which is 1 way. Hence, I have: $${{5}\choose{2}} \cdot 6 {{3}\choose{3}} \cdot 5 \cdot 4 \cdot 3 = 3600 $$ However, I couldn't figure out why I got two pairs wrong. First, I pick 2 places for the first pair, then its rank. Next, 2 places for the second pair, and its rank. Since there is only 1 place left, I pick the rank for the last dice. $${{5}\choose{2}} \cdot 6 {{3}\choose{2}} \cdot 5 \cdot 4 \cdot 3 = 3600$$ why the correct answer is 1800, which means I need to divide by a factor of 2. I guess that might be the order of two pairs can be switched, but I wonder is there a better way to count it? I'm so confused :(! Any idea?

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up vote 4 down vote accepted

You’ve correctly identified the mistake: you’ve counted each hand twice, because the pairs can be chosen in either order. For instance, you’ve counted the hand $11223$ once with $22$ as the first pair and $33$ as the second pair, and once again with $33$ as the first pair and $22$ as the second pair.

Here’s a way to count that avoids that problem. First pick the two denominations that will be pairs; this can be done in $\binom62$ ways. Then pick the remaining denomination; this can be done in $4$ ways. Now choose which of the $5$ dice will show the singleton; this can be done in $5$ ways. Finally, choose which $2$ of the remaining $4$ dice will show the smaller of the two pairs; this can be done in $\binom42$ ways. The total is then $\binom62\cdot4\cdot5\cdot\binom42=1800$ ways. The key to avoiding the double counting is to choose the positions for a specific pair. Once you know where the smaller pair and the singleton are, you automatically know where the larger pair is: there’s nothing to choose.

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Thanks a lot. I saw my problem now. In fact, I was confused this problem with counting poker hand. – Chan Sep 14 '11 at 9:57

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