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The problem is:

How many are ways to deal a deck of 52 cards to 4 players, and every player has at least one card?

The answer with inclusion-exclusion principle is: $$4^{52}-4\cdot 3^{52}+6\cdot 2^{52}-4 $$

But I'm wondering, why isn't it equal to the number of solutions: $$a+b+c+d=52, \ a,b,c,d\in\mathbb{N}, a,b,c,d>0$$ which we can calculate by finding coefficient before $x^{52}$ in expansion to series this function: $$\frac{x^4}{(1-x)^4}=\left(\sum_{n\ge 1}x^n\right)^4$$ which is quite easy? The result is completely different(extremely less). What we don't count this way, what we are missing? Can this approach be fixed?

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1 Answer 1

up vote 1 down vote accepted

With the first solution you compute the number of different deals, whereas with the second, you only compute the number of different size of deals (what number of cards each player has). So you must find a much bigger number with the first solution.

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There is no simple "magic number" by which you can multiply the answer from the second approach to get the right answer. This is because for example the number of deals of size $1$-$1$-$2$-$48$ is quite different from the number of deals of size $10$-$10$-$10$-$22$. –  André Nicolas Aug 1 '12 at 14:17
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