How many five-card hands contain exactly one pair of the same rank and three other cards of different ranks?
This is my thought process:
I first multiply it by $13 \binom{4}{2}$, since there are $13$ possible rank pairs we can create and then multiply by $12\cdot11\cdot10$ for each different rank we choose. Meaning the total number of combinations is $$13 \cdot \dbinom{4}{2} \cdot 12 \cdot 11 \cdot 10$$
Is this right?