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I found an interesting solution to the combinatorial question of "How many poker hands have exactly two pairs?" and I cannot figure out (or find) the reasoning of the solution. The answer I found in the textbook I'm studying from is:

${\huge{\frac{\frac{52 \times 3}{2}\times \frac{48 \times 3}{2}}{2}\times44}}$

Can someone explain the logic behind this solution?

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First, assume you choose the cards in order of pair 1, pair 2, non pair card.

The first card in pair 1 can be any of 52 cards. The second card in pair 2 can be one of 3 cards (the other cards of the same number). The first card of pair 2 can be one of the 48 cards which are not of the number in pair 1. The second card of pair 2 is one of the 3 cards of the same number as the first card in the pair. The last card in the hand is not of the number in the first pair or second pair, so there are 52-8 = 44 cards which you can select from.

The division by 2's in the numerator account for ignoring order in each pair and ignoring the order the pairs were selected in.

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