Suppose we had a standard $52$ card deck. How many different full house sets are there if we take suits into consideration? For example:
$\rm(8\clubsuit, 8\heartsuit, Q\heartsuit, Q\spadesuit, Q\diamondsuit)$ and $\rm(8\heartsuit, 8\spadesuit, Q\clubsuit, Q\spadesuit, Q\diamondsuit)$ are different full house sets.
My original intuition is to multiply all different possibilities for a $2$ card rank, $2$ card suit, $3$ card rank, $3$ card suit.
Leaving me with ${13 \choose 2} \times {4 \choose 2} \times {13\choose 3} \times {4 \choose 2}$
Choose the 3-card rank, 3 cards from that rank, 2-card rank, and 2 cards from that rank: $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}=13 \cdot 4 \cdot 12 \cdot 6 = 3744$$