I'm working through "Schaum's" notes on probability in preparation for a (non-academic) exam. I was asked to calculate the probability that drawing 5 cards from a standard deck will result in a full house. My answer disagrees with theirs by a factor of 6, which makes me think I'm double counting based on order (or something like that).
The strategy: count the number of admissible hands, divide by the number of possible hands (52-choose-5). A hand is "admissible" if it is a full house: if it is made up of 3 cards from one "rank" and 2 cards from another "rank".
I guess I am messing up the counting:
For each of the 13 ranks, there are (4-choose-3) ways to make a triple. For each of the remaining 12 ranks, there are (4-choose-2) ways to make a pair.
Any triple can be joined with any of these pairs, so that these quantities multiply: