When to use ${13 \choose 1}\times{12\choose 2}$ and not ${13\choose 3}$ It's really a conceptual problem and I am looking for a good way to get the intuition right more often than chance. Typical occurence:

count the number of poker hands containing exactly three cards of the
  same rank (with the other two cards different ranks)

Here the answer is ${13\choose 1}{4\choose 3}{12\choose 2}{4\choose 1}^2$ or put another way ${13\choose 1}{12\choose 2}{4\choose 3}{4\choose 1}^2$. 
My problem is with the the first two terms. They stand for "picking 1 rank out of the 13 possible and then 2 ranks out of the remaining 12". But how is this different from "picking 3 ranks out of the possible 13" which would be  ${13\choose 3}$. How to know when to use the former and not the latter. 
 A: It is true that choosing 1 from 13 and then 2 from 12 is the same as choosing 3 from 13.  But the issue is that each choice of rank for the three matching cards must be paired with every other choice for the remaining two ranks.  The fact that $\binom{13}{1}$ and $\binom{12}{2}$ are being multiplied is important.  Merely looking at $\binom{13}{3}$ wouldn't account for this multiplicity.  If on the other hand you were to multiply $\binom{13}{3}$ by 3, then you'd get back to the right answer, because then every rank occurring in each three card combination would have a turn at being three of a kind.
To illustrate simplify the problem and suppose there are only three ranks and denote them $\{1, 2, 3 \}$.  Choosing the three ranks for your hand is just taking the whole set and would be like counting $\{1, 1, 1, 2, 3 \}$ as the total number of combinations.  If instead you multiply this by three you would be counting each of $\{1, 1, 1, 2, 3 \}$, $\{1, 2, 2, 2, 3 \}$ and $\{1, 2, 3, 3, 3 \}$.  This is the same as first choosing the number to be matched three times and then multiplying by the number of ways of choosing the remaining two ranks, which happens to be one here.
