Count the number of 5 cards such that there's exactly 2 suits

Question

Count the number of 5 cards such that there's exactly 2 suits

Suppose we draw five cards from a standard deck of 52 cards. I want to count the number of ways I can draw five cards such that the hand contains exactly 2 suits.

Here's my intuition:
There are two cases, one case involves a single card with a different suit from the other 4, and the other case involves two cards with the same suit, and the other three have a different suit.

Case 1: ${4 \choose 1} {13 \choose 1} \cdot {3 \choose 1} {13 \choose 4}$

Case 2: ${4 \choose 1} {13 \choose 2} \cdot {3 \choose 1} {13 \choose 3}$

Am I correct with the cases? I'm confused on the coefficients for choosing the suits for the first group of cards since it will limit the number of suits to choose for the next group of cards. Should I multiply each case by $2$ ?

Answer 1

Yes, this is correct without multiplying by 2. You can check by computing a different way. Choose two suits, choose all five cards from these two suits, and subtract the ways that yield only one suit: $$\binom{4}{2}\left(\binom{26}{5}-\binom{2}{1}\binom{13}{5}\right)=379236$$

Answer 2

Your cases 1 and 2 are correct, so the total number of hands satisfying the property in question is $\binom{4}{1} \binom{13}{1} \binom{3}{1} \binom{13}{4} + \binom{4}{1} \binom{13}{2} \binom{3}{1} \binom{13}{3}$.

There is no need to multiply anything by 2. It is true that the selections of the suits of the large and small groups are not independent, but whatever you choose for the first one (in 4 possible ways), there will be 3 choices for the second one.