How many 5 card poker hands have at least one card from each suit, but no two matching values?

Question

How many 5 card poker hands have at least one card from each suit, but no two matching values?

There are 4 possible suits for the first suit, of which we have 13 choices, then 3 possible suits for the second, of which there are 12 choices of values, then 2 possible choices for the third suit of which there are 11 possible choices, and then 1 possible suit for the fourth of which there are 10 possible choices. Finally for each of these choices, there are 36 possible cards for the fifth card. So the answer would be $4!(13)(12)(11)(10)(36)=14,826,240$.

However, I don't think thisis right, because I believe I'm double counting many hands.

Answer 1

Another way: you can select the repeated suit in $\binom{4}{1}$ ways, the values for the repeat suit in $\binom{13}{2}$ ways, and asign the rest of the values to the remaining suits (in order) in $11 \cdot 10 \cdot 9$ ways, for a total of $$ \binom{4}{1} \cdot \binom{13}{2} \cdot 11 \cdot 10 \cdot 9 = 308880 $$ As a general rule, it is best to reduce such problems to a sequence of decisions, counting sequences is easy (and the risk of over/under counting is less).

Answer 2

You're right to be suspicious. You can get the same hand dealt in a different order, but you would count it differently. How many different ways can you be dealt the hand you've described?

Answer 3

There are $4$ choices to select the one suit that has two cards. There are $13\choose 5$ ways to choose the values. There are $5\cdot 4\cdot 3$ ways to assign the suits to the values. That gives us $4\cdot {13\choose 5}\cdot 5\cdot4\cdot 3=308880$ such hands in total.