How many hands of five cards contain cards from exactly three suits?

Question

I have been trying to solve this but I stink at this kind of math. What I have is

(4 choose 3)(13 choose 2)(13 choose 2)(13 choose 1)=316,368.

Is this answer right or wrong? Thank you!

Answer 1

You are almost there, but you also need to account for the choice of which suit gets only 1 card (or which suit gets 3 cards in the case of 3,1,1).

Here's an alternative approach. There are $\binom{52}{5}$ five-card hands in total.

• $\binom{4}{1}\binom{13}{5}$ contain exactly 1 suit.
• $\binom{4}{2}\sum_{k=1}^4 \binom{13}{k}\binom{13}{5-k}$ contain exactly 2 suits.
• $\binom{4}{4}\binom{4}{1}\binom{13}{2}\binom{13}{1}^3$ contain exactly 4 suits.

Now subtract these three counts from $\binom{52}{5}$, yielding $1529112$.

Answer 2

A five-card hand will contain exactly three suits when it contains either of:

• (Two doubles and a single) Two from the thirteen kinds in each of two from the four suits, and one from the thirteen kinds in one from the two remaining suits, or
• (Two singles and a tripple) One from the thirteen kinds in each of two from the four suits, and three from thirteen kinds in one from the two remaining suits.