Combinatorics - $5$ cards, $4$ different suits I have the following question:

In a deck of $52$ cards with $4$ suits ($13$ of each), how many different ways are there to choose $5$ different cards such that every suit appears at least once.

the correct answer is:

$4×13^3×{13\choose 2}=685464$

My question is, why is the following wrong:

$\frac{52*39*26*13*48}{5!}$

As $52$ is the first card, then we want $39$ as we don't want from the first suit, then $26$, and $13$, and then $48$ as for the last one we can choose again any suit.
Then divide by $5!$ as we don't care about the order.
Now I know this is wrong obviously as wee don't even get an integer.... the interesting thing is that when dividing by $2*4!$ as in $\frac{52*39*26*13*48}{2*4!}$ we get the same result as above, and also when just multiplying $52*39*26*13$ we get the correct result...
I can't figure out where did I go wrong,
thanks for the help!
 A: Here is an alternative solution.

Use inclusion/exclusion principle:


*

*Include the number of combinations with at most $\color\red4$ suits: $\binom{4}{\color\red4}\cdot\binom{13\cdot\color\red4}{5}$

*Exclude the number of combinations with at most $\color\red3$ suits: $\binom{4}{\color\red3}\cdot\binom{13\cdot\color\red3}{5}$

*Include the number of combinations with at most $\color\red2$ suits: $\binom{4}{\color\red2}\cdot\binom{13\cdot\color\red2}{5}$

*Exclude the number of combinations with at most $\color\red1$ suits: $\binom{4}{\color\red1}\cdot\binom{13\cdot\color\red1}{5}$


Hence the total number of combinations is:
$$\sum\limits_{n=0}^{3}(-1)^{n}\cdot\binom{4}{4-n}\cdot\binom{13(4-n)}{5}=685464$$
A: Let the $4$ suits be $A,B,C,D$.
Since you have to choose $5$ cards and a card from each suite must appear at least once,so the following possibilities occur :


*

*$2$  from $A$ and remaining three from $B,C,D$ one from each.

*$2$ from $B$ and remaining three from $A,C,D$ one from each.

*$2$ from $C$ and remaining three from $A,B,D$ one from each.

*$2$ from $D$ and remaining three from $A,B,C$ one from each.


Hence number of ways in which any one can occur is $13C_2\times 13C_1\times 13 C_1\times 13 C_1$
Now any suite can be chosen in $4C_1$ ways.
Hence number of ways is $4C_1\times 13C_2\times 13C_1\times 13 C_1\times 13 C_1=4\times 13C_2\times 13^3$ 
A: First thing $\frac{52*39*26*13*48}{4*4!}$ and $52*39*26*13$ are not same.
Second - After picking 5 cards we have 2 cards of same suit. These 2 are further arrange in 2 ways. Hope you can understand now.
A: The problem lies behind your selection method and $5!$ at the denominator. 
You've assumed $52$ choices for the first card, $39$ choices for the second, $26$ choices for the third, $13$ choices for the fourth, and $48$ choices for the last card. For now, let's put the $5!$ denominator aside and deal with ordered combinations. Now, for example, one combination could be this (notice the hearts) $$5\heartsuit\;\; 2\diamondsuit\;\; 2\clubsuit\;\; 2\spadesuit\;\; 8\heartsuit$$ which means the $5\heartsuit$ was selected as the first card and the $8\heartsuit$ as the last. Of course you've counted many other combinations of the same cards, like this $$2\diamondsuit\;\; 2\clubsuit\;\; 5\heartsuit\;\; 2\spadesuit\;\; 8\heartsuit$$
But the problem is you've never counted combinations like this
$$5\heartsuit\;\; 8\heartsuit\;\; 2\diamondsuit\;\; 2\clubsuit\;\; 2\spadesuit$$ where one of the two same suit cards is not selected as the last one. The reason is your selection method always selects $4$ distinct suits first and then a repeated one for the last position. Therefore, the total combinations of these cards in your selections is not $5!$. That's why your answer was not even an integer. Actually you only have $2\times4!$ combinations; $4!$ ways to rearrange the $4$ first distinct suits, and $2$ ways to change the repeated cards ($5\heartsuit$ and $8\heartsuit$). So the correct answer is: $$\frac{52\times39\times26\times13\times48}{2\times4!}$$
Hope it helps.
