How many ways are there to distribute a deck of $52$ cards to $4$ players?

Question

This is a homework problem that me and my friend got confused on. It says:

"How many ways are there to distribute a deck of $52$ cards to $4$ players?"

The teacher clarified that each player does not have to have $13$ cards each, meaning $1$ player can have $52$ cards and the others have $0$, $2$ players have $26$, $26$, etc.

So my friend said that the answer should be $4^{52}$ because you have $52$ cards, and each card can be chosen by $1$ of the $4$ players. So you have $4$ options for the first card, $4$ options for the second card, $4$ options for the $52$nd card.

On the other hand, I said that the answer should be $C(4, 52)$, in other words, $\frac{55!}{52!3!}$, because you have $4$ players total with $52$ cards to distribute. Since the problem said the ways you can distribute the deck, the cards itself do not matter. So you are solving for the number of ways you can have $x_1 + x_2 + x_3 + x_4 = 52$ (with $x_1$ to $x_4$ being each player).

I really don't know which method is correct. Both methods make sense to me but clearly there can only be 1 correct answer.

Answer 1

Your friends method and the Stars and Bars method have one fatal difference between them.

The Stars and Bars method doesn't care which cards goes to whom; only the number of cards each player has is what distinguishes the different possibilities. Think of it as having $52$ cards but they are all blank.

Your friends method is actually correct in this case. What we want is the number of ways to distribute $52$ different cards between four people. For example, let player 1 have $51$ cards and player 2 have the last card while players 3 and 4 don't get any cards. Stars and bars would only count this outcome once, but that isn't correct. Maybe player 2 has the Ace of Spades, or the King of Clubs, or the $9$ of diamonds, etc. We need to consider these different outcomes when figuring out the total number of outcomes, which is $4^{52}$

Answer 2

Your friend is correct. What matters here is who receives which card.

There are four possible recipients for each of the $52$ cards, so there are $4^{52}$ ways to distribute the cards to four people without restriction.

What you counted is the number of ways of distributing $52$ identical index cards to four people. If the cards are indistinguishable, what matters is how many cards each person receives. If we let $x_i$ be the number of cards received by the $i$th person, then the number of ways of distributing $52$ indistinguishable index cards is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 = 52$$ in the nonnegative integers. As you seem to be aware, this is equal to the number of ways of inserting three addition signs in a row of $52$ ones, which can be done in $$\binom{52 + 4 - 1}{4 - 1} = \binom{55}{3}$$ since we must choose which three of the $55$ positions required for $52$ ones and three addition signs will be filled with addition signs.