What is the probability of partitioning a deck in to 13 subsets of 4 cards each out of the deck of 52 cards having all different ranks? 
What is the probability of partitioning a deck into 13 subsets of 4 cards each out of the deck of 52 cards having all different ranks?

My solution is that since there are $_{52}C_4$ ways to pick 4 cards out of the $52$ card stack. and since we need $13$ hands we multiply by $13$. so $13\cdot (_{52}C_4)$ and then there are $_{13}C_4$ ranks and each rank could be $1$ out of $4$ possible suits. So it will be $(_{13}C_4)\cdot4^4$. Therefore everything all together will be: $$\frac{4^4 \cdot(_{13}C_4) }{ 13\cdot(_{52}C_4)}$$
Can someone confirm if this is the correct approach?
 A: For problems that ask for just one hand of four cards out of $52$, it often makes sense to say that there are $\binom{52}4$ different hands you are equally likely to receive.
I say "often", because in order for this to be useful you need to be able to count the "good" hands in a similar way, so that each "good" hand that is counted once in the denominator (the same four cards could be set into $24$ different sequences, but we still only count them once) is also counted exactly once in the numerator.
Now suppose you are dealt just two hands of four cards each.
The first hand might be any of the $\binom{52}4$ different hands we counted in the first paragraph.
But if the first hand is (for example) 
$\spadesuit 4, \spadesuit \mathrm J, \heartsuit 3, \clubsuit 10,$
then you know that none of these four cards can be in the second hand.
Only $48$ cards remain from which to choose the second hand,
so there are $\binom{48}4$ such hands.
However, since we can pick any of $\binom{52}4$ hands for the first hand, and then for each of those we can pick the second hand in $\binom{48}4$ ways,
there are altogether
$$\binom{52}4 \times \binom{48}4 $$
ways to choose just two hands of four cards each.
That's already a lot more than $13 \times \binom{52}4.$
And you still have eleven more hands to choose.
I'm assuming the $13$ hands are all chosen from the deck without replacement, that is, we are really just partitioning the deck into $13$ subsets of four cards each.
If each hand were put back in the deck and the deck were shuffled before dealing the next hand, then the number of possible ways to deal the first two hands would be
$\binom{52}4^2,$ which is even larger than for two hands dealt without replacement.
Now, if you have some way of counting the hands so that you do not distinguish the case where you get
$\spadesuit 4, \spadesuit \mathrm J, \heartsuit 3, \clubsuit 10$
in the first hand and 
$\spadesuit 9,  \heartsuit \mathrm K, \diamondsuit 6, \diamondsuit 7$
in the second hand from the case where you get
$\spadesuit 9,  \heartsuit \mathrm K, \diamondsuit 6, \diamondsuit 7$
in the first hand and 
$\spadesuit 4, \spadesuit \mathrm J, \heartsuit 3, \clubsuit 10$
in the second hand,
then you have only
$\frac12 \times \binom{52}4 \times \binom{48}4 $
different ways to get two hands, but that's still a lot more than
$13 \times \binom{52}4.$
Meanwhile, for the numerator you've represented the number of ways you can get four different ranks in one hand. What about the other twelve hands?

I think we are both assuming that the problem statement is asking for the probability that in each of the $13$ hands there are no two cards of the same rank.
The problem statement is not completely clear about this.
It might be a good idea to ask the person who posed the problem, if you can,
to make sure this is the correct interpretation.
