Counting the number of hands which contain a 3 of a kind

Question

When trying to find the number of three of a kind hands in a standard deck of cards such that there are three of one number, one of another, and one of one more number, I got what I believe to be the correct answer of $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}$$ possible combinations. The part of this that confuses me is why is the answer not $$\binom{13}{3}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ Why cannot I not account for all of my number choices within the first choose? Why do I need to differentiate between the 3 of a kind and the 2 number not in the 3 of a kind? Why can I not combine the $\binom{13}{1}$ and the $\binom{12}{2}$?

Answer 1

to do it your way (which is perfectly valid) you have to remember to choose which of the $3$ selected ranks will be the one with the triplet. So you should have gotten $$ \binom{13}{3}\color{red}{\binom{3}{1}}\binom{4}{3}\binom{4}{1}\binom{4}{1} $$ which is equal to the given answer.