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Attempt:

  1. If the cards are distinct?

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

  1. 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?

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    Not clear what you mean in 1 and 2. If the 4 persons take different cards or they can take the same? I think you need to clarify the question. – 2017-02-01
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    in 1. it's assumed that all 52 cards are different while in 2. all 52 cards are the same – 2017-02-01
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    Why wouldn't it still be $4^{52}$? I mean, you start with dealing, for instance, the ace of spades. There are four people who might get the card. Then on to the king of spaces. Again four options. And so on. (I must admit, though, that I'm not entirely sure where $\binom{51}{3}$ comes from.) – 2017-02-01
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    In my book it gave us the general form for distributing distinct items to distinct people/things as $\binom{m-1}{n-1}$ but the condition to use this is that each person be guaranteed at least 1 card or nonempty. Or is the $4^52$ for part 2 incorrect, I'm not really entirely sure on both. – 2017-02-01
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    I'm still thinking about 1 and $4^{52}$ – 2017-02-01
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    what if you had 2 cards and 2 people? Why would the answer be $\binom{1}{1}$ What are the rules for distributing to give the answer of 1? I can only see that id for non-distinct and they both have to receive 1 – 2017-02-01
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    I misread the part in the book. This deals with identical cards and distinct people. or identical balls and distinct boxes to put the ball in. So this would mean 2's answer if everyone had at least 1 card would be $\binom{51}{3}$ but this still wouldn't matter if people were allowed to have 0 cards. – 2017-02-01
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    in 1, if people can not have zero cards, you have to subtract all the deals that give zero cards to anyone, which is an exclusion, so working out that A has zero cards and multiplying by 4 does not quite work, it double counts cases where A and B have zero, then there are the cases where 3 have zero - Dr Hab has worked it out already – 2017-02-01

2 Answers 2

2

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

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    hmmm I think I follow but wouldn't it be $\binom{55}{4}$ not 3? – 2017-02-01
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    if you were splitting 5 identical cards to 2 people, consider XOXXXX moving the O to all possible positions generates all possible shares - but there is 1 partition marker to split into 2 - splitting like this requires n-1 partitions - see 'stars and bars' – 2017-02-01
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    ah i see. I'm actually not 100% on the question if we can deal 0 cards, the question seems ambiguous, but I'm only studying right now so thanks for explaining both sides. – 2017-02-01
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    if 0 was allowed wouldn't the answer be $\binom{55}{4}$ and if not allowed, the answer would be $\binom{51}{3}$ no? – 2017-02-01
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    why $\binom{55}{4}$ would that mean there were 4 dividing cards, dividing into 5 groups - is my reckoning – 2017-02-01
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    I figured out the correct way the equation works. It's $\binom{52+4-1}{52}$ which is the same as $\binom{55}{3}$ – 2017-02-01
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    @Cato In case of $4^{52}$ there is a possibility of getting string $52525252$ interpreted as person $1$ gets $52$nd card and so on..which is not possible in practice. – 2017-11-24
2

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 $\{1,\dots,52\}\to\{1,2,3,4\}$.