Ways to distribute $52$ cards to $4$ distinct people Attempt:


*

*If the cards are distinct? 


If cards are distinct I have no idea how to proceed this.


*If the cards are identical?


I could treat this as if there are $4$ boxes and $52$ identical balls which would have $4^{52}$ ways to distribute the balls/cards. 
However, I am not sure how to proceed for if the cards are distinct. If people weren't allowed to have $0$ cards, I know that it would be $\displaystyle\binom{51}{3}$. Do I just take that and then just add all the ways where people end up with $0$ cards?
 A: 1 is $4^{52}$ - assuming you can have zero cards
2 is found by considering 3 partitions mixed withing the cards - moving the partitions generates the different allocations of cards to 4 people, since there are 3 partitions to move and 52 cards, making total of 55, the number of arrangements is
$_{55}C_{3}$
edit - sorry that assumes you can have zero cards - if you can't have zero cards then you can give every one 1 card (1 way of doing this) then distribute the remaining 48 cards as above, except this time it will be $_{51}C_{3}$ - your answer - so yes I agree with your answer on 2
A: This answers 1) under the extra condition that each person gets at least one card. For the other $3$ cases see the answer of Cato.
Without that condition there are $4^{52}$ possibilities. 
Give the persons an index. For $i=1,2,3,4$ let $U_i$ denote the subcollection of possibilities where person $i$ does not receive a card.
Then to be found is: $$4^{52}-|U_1\cup U_2\cup U_3\cup U_4|$$
With inclusion/exclusion and symmetry we can find: 
$$|U_1\cup U_2\cup U_3\cup U_4|=4|U_1|-6|U_1\cap U_2|+4|U_1\cap U_2\cap U_3|$$
This results in:$$4^{52}-4\times3^{52}+6\times2^{52}-4$$possibilities.
You can think of it as the number of surjections $\{1,\dots,52\}\to\{1,2,3,4\}$.
