Card problem - combinations? Deal a pokerhand (5 cards) with 5 different ranks and 4 different suits.
EG:
2♠ (spades)
3♥ (hearts)
4♦ (diamonds)
5♣ (clubs)
6♠ (spades)

I have been told that this is the solution:
$\binom{13}{5}\binom{4}{1}\binom{5}{2}\times 3\times 2$
I'm trying to solve it like this, and I really cannot see why it wouldn't work (it is wrong):
$\binom{13}{1}\times 4\binom{12}{1}\times 3\binom{11}{1}\times 2\binom{10}{1}\binom{9}{1}\times 4$
Explaining how I was thinking:


*

*Choose 1 card out of 13 - 4 possible colors. $\binom{13}{1}\times 4$.

*Choose 1 card out of 12 - 3 possible colors. $\binom{12}{1}\times 3$.

*Choose 1 card out of 11 - 2 possible colors. $\binom{11}{1}\times 3$.

*Choose 1 card out of 10 - 1 possible color. $\binom{10}{1}\times 1$.

*Now I have 4 cards, 4 different suits and 4 different ranks.

*Choose 1 card out of 9 - 4 possible colors. $\binom{9}{1}\times 4$.


Then just multiply everything together, why doesn't this work..?
 A: *

*Your answer cares about the order you receive cards in; you count as different the two sets 

2♠ 3♥ 4♦ 5♣ 6♠

and

3♥ 2♠ 4♦ 5♣ 6♠

so you are not correctly enumerating only different combinations. You count some things too many times.

*If you had correctly counted all permutations, you would just be able to divide by $5!$ at the end and still get the right answer - however, you haven't quite counted all permutations either, because you have insisted that you get the duplicate suit as the last card. In fact, for every possible combination, your approach lists it exactly 48 times.
Why? Look at the example. You could choose either the 2♠ or 6♠ for your special last card (a factor of $2$). You could also then have the first four cards in any order, giving $4! = 24$ further choices.
So it is possible to obtain the answer from your result by dividing by 48.

Edit: Of course, this isn't the easiest way to get the answer. Their form of the answer is reasoned as follows:


*

*Each combination has exactly $5$ of the $13$ ranks, hence a $13 \choose 5$ factor.

*Exactly $1$ of the $4$ suits appears twice, hence $4 \choose 1$.

*Exactly $2$ of the $5$ cards are this special suit, hence $5 \choose 2$.

*The remaining $3$ suits and $3$ cards can be matched up in exactly $3! = 3\times 2 \times 1$ ways, hence the last term.

A: You over count. Your method considers selecting $2\heartsuit,3\spadesuit,4\clubsuit,5\diamondsuit,$ then $6\spadesuit$ for the extra as distinct from $2\heartsuit,6\spadesuit,4\clubsuit,5\diamondsuit,$ then $3\spadesuit$ as the extra.

To have five different ranks and four different suits you must have two cards of one suit and one each of the other three.


*

*select 5 from 13 ranks.

*select 1 from 4 suits as the doubled suit.

*arrange 4 singletons and 1 double; matching suits to ranks by:


*

*selecting the two cards to have the doubled suit

*arranging the single suits among the remaining three




 $${^{13}\mathrm C_4}{^{4}\mathrm C_1}{^{5}\mathrm C_{2}}3!$$

