I have seen some answers to the question as $$\frac{\displaystyle\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2}{\displaystyle\binom{52}{5}} $$
Why don't we use $$\frac{\displaystyle\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{11}{1}\binom{4}{1}^2}{\displaystyle\binom{52}{5}} $$ Which makes sense to me. There are 13 ranks and you choose 1. Of the one chosen there are 4 cards of which you choose 3. Now there are 12 ranks of which one is chosen. Now 4 choose 1. Leaving for the final card 11 ranks of which one is chosen. And choose 1 of 4. Divide all this by the number hands
The explanation that is given is that (aaabc)=(aaacb). But why? IF c came first, THEN call it b.
Explaining the correct numerator.
$$13\times\binom{4}{3}\times\binom{12}{2}\times 4^2=54{.}912$$
This because
you have 13 ways to choose the set
$\binom{4}{3}=4$ ways to choose the suits of your set
you have $\binom{12}{2}$ ways to choose the remaining (and different) cards where order does not matter.
you have $4^2$ ways to choose the suits for the 2 remaining cards