How many different 5 cards hands are there containing all the four suits? Given a deck of $52$ cards, I'm trying to calculate how many hands there are containing all four suits. The result I'm supposed to get is $685464$, but I'm not sure about the method to be used.
My attempt

*

*I have found this related question, so I understand that the answer will be the number of suits to be chosen (i.e., ${n \choose 1}$, with $n=4,3,2,1$) times the number of cards to be chosen (${13 \choose 2}$ for the repeated suit and ${13 \choose 1}$  for the non-repeated ones), so

$$
N={4 \choose 1}{13 \choose 2}·{3 \choose 1}{13 \choose 1}·{2 \choose 1}{13 \choose 1}·{1 \choose 1}{13 \choose 1} = 4112784
$$
But it is 6 times bigger than the expected result.


*I have also thought about computing the number as the product of the number of options to pick a card from a suit, ${13 \choose 1}$, four times, and the number of options of picking a card out of the remaining $48$: ${48 \choose 1}$, so

$$N={13 \choose 1}^4{48 \choose 1} = 1370928$$
But it is now 2 times bigger than the expected result.
Why do these approaches fail? What would be the correct way?
 A: This was slightly too long for a comment, so it became an answer.
(1) Consider the suits to be in a fixed order, for instance $\clubsuit$, $\diamondsuit$, $\heartsuit$, $\spadesuit$. Fix a $5$-cards poker hand with all four suits, and rearrange it so that we have first the suit with two cards, then the remaining three suits in the above fixed order. The hand is of the shape $AABCD$, when only suits are shown. This uniquely determines the hand, and counting we get $4=\binom 41$ possibilities for the choice of $A$ among $\clubsuit$, $\diamondsuit$, $\heartsuit$, $\spadesuit$, then $\binom{13}2$ possibilities for the cards $AA$, then each $13=\binom{13}1$ for the $B$, the $C$, and the $D$ card. The obtained total number is thus:
$$
\binom 41\cdot \binom{13}2\cdot \binom{13}1^3=685464\ .
$$
The value in $(1)$ in the OP also has the factors $\binom 31$ and $\binom 21$, using them would count each $AABCD$ with the above convention also realized in the six cases of a permutation of $BCD$:
AABCD
AABDC
AACBD
AACDB
AADBC
AADCB

(2) Let us refine this second way of counting, so that we definitively do not count a hand twice. Fix a hand. Reorder it as follows. The first four cards are in the fixed order of the suits. For one of them we have a choice. Alice will take the bigger card among the two. Then at the last place Alice puts the lower card from this doubly represented suit. So Alice gets a hand of the shape $\clubsuit\diamondsuit\heartsuit\spadesuit X$. But $X$ is not freely chosen among the remained $48$ cards, it has to be lower than the card $A$ already chosen in the same suit.
Bob uses the similar scheme, but takes first the lower card among the two, and puts at the end the bigger card in the same suit.
If we use the counting from (2) for the shape $\clubsuit\diamondsuit\heartsuit\spadesuit X$
$$
\binom{13}1^4\cdot\binom{48}1\ ,
$$
with $X$ arbitrary among the remained $48$ cards, remained after the choice of the first four $\clubsuit\diamondsuit\heartsuit\spadesuit$,
then we get a hand, which - as it is -  ordered either in the sense of Alice, or in the sense of Bob. Each hand is thus counted twice. We have to divide by two to get the right answer.
Later EDIT: I will give an example for (2), since often words do not determine exactly what they were supposed to describe. The order of operations may count, the ordering may count, etc. - and combining the pieces of the puzzle is simpler checked with an example. Let us consider the following "unordered" hand:
$$
Q\heartsuit\ ,\
9\diamondsuit\ ,\ 
K\heartsuit\ ,\
10\spadesuit\ ,\ 
8\clubsuit\ .
$$
In the OP it is counted somewhere among the $N=13^4\cdot 48$ cases, let us see where... For this let us pre-sort the hand, and impose after pre-sorting for the first four cards, (the ones counted via the $13^4$ factor,) the order fixed above. But (even) with this restriction, we do not have a determined sorted hand, since we (can) count the above sample hand twice, explicitly in the following two orders:
$$
\begin{aligned}
H_1 &= 
8\clubsuit\ ,\ 
9\diamondsuit\ ,\ 
\color{red}{Q\heartsuit}\ ,\
10\spadesuit\ ,\ &&\text{ and remaining card }K\heartsuit\ ,
\\
H_2 &= 
8\clubsuit\ ,\ 
9\diamondsuit\ ,\ 
K\heartsuit\ ,\
10\spadesuit\ ,\ &&\text{ and remaining card }\color{red}{Q\heartsuit}\ ,
\ .
\end{aligned}
$$
This is the reason for introducing Alice and Bob above. They have each a sorting strategy that determines the hand. Alice would sort the above sample hand obtaining $H_2$. Bob will sort it obtaining $H_1$. It is "hard" to count the hands seen by Alice separatedly, or the ones of Bob. But we take them together, and count the total easily, it is $13^4$ for the possibilities of the first $4$ cards in the four different suits, ordered by the suit order, times $48$ for the possibilities for a final fifth card. We know that we get the doubled number of possible hands.
