+101 You are dealt cards from a standard deck of 52 cards.
How many ways can you be dealt the cards so that they contain two cards of one rank, two cards of another rank, and a fifth card of a third rank? We say that such a hand has two pairs. For example, the hand QQ225 has two pairs. (Assume that the order of the cards does not matter.)
I thought that there would be 13 choices for the first card, 1 for the second, 11 for the third, 1 for the fourth, and 9 for the fifth. Multiply that and get 1287.
However, my answer is showing as wrong. Could someone please explain?
