Drawing 3 cards of same suit and 2 of a different suit What is the probability of having 3 cards of the same suit and $2$ cards of the same suit (but a different suit than the first three) in a $5$ card hand from a standard $52$ card deck? 
What is the difference between solving this problem using approach (1) and (2). Assuming we divide both by ${52 \choose 5}$
$(1)$
$$
{4 \choose 1}\cdot{13 \choose 3}\cdot{3 \choose 1}\cdot{13 \choose 2}
$$
$(2)$
$$
{4 \choose 2}\cdot{13 \choose 3}\cdot{13 \choose 2}
$$
Why can't I just pick the two suits all at once? Why do they have to be separated out? In the first approach I don't have the problem of duplicates. A {10_Hearts, J_Hearts, K_Hearts, 3_Spades, Q_Spades} doesn't equal {3_Spades, Q_Spades, X_spades, 10_Hearts, J_Hearts}
 A: Revised to match corrected question: If you pick the two suits at once, you don’t distinguish the case of (say) $3$ hearts and $2$ clubs from the case of $3$ clubs and $2$ hearts. Thus, you ought to get exactly half of the right answer, and indeed you do:
$$\binom42=6=\frac12\binom41\binom31\;.$$
If you want to start out by choosing the two suits, you can, but then you have to choose which is to get $3$ cards: all that can be done in $\binom42\binom21=12$ ways. That’s exactly what you get if you first choose the $3$ card suit and then the $2$-card suit from the $3$ remaining suits: $\binom41\binom31=12$. And you still have to choose the cards from the suits, so you still get factors of $\binom{13}3$ and $\binom{13}2$.
A: Select the suit with 2 cards (1 out of 4), select their values (2 out of 13), then select the suit with 3 cards (1 out of the remaining 3) and then the 3 values (3 out of 13) for a total of
$$
\binom{4}{1} \cdot \binom{13}{2} \cdot \binom{3}{1} \cdot \binom{13}{3} = 267696
$$
