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!

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.
