Implement genarating function for a combinatorial question

Question

I've tried to solve the following question by using generating functions.I know It's pretty simple to do it with Inclusion–exclusion principle but I insist to solve it with other technique as well.

Q: how many combination to spread 10 balls into 5 boxes when in each box there is at least one ball.

A: $ (x+x^2+x^3+x^4...)^5 = x^5(1+x+x^2+x^3...)^5 = x^5 \frac{1}{(1-x)^5} = x^5 \sum_{i=0}^\infty \binom{n+5-1}{5-1}x^n$

I want the coefficient of $x^{10}$ so $n=5$ which is $\binom{9}{4} = 126$

sadly it isn't the same answer like the Inclusion–exclusion principle, i guess my series isn't right but I can't see why.

Answer 1

This is small enough to check by hand which of your results (if either) is correct.

$6$: $5$ arrangements

$5$, $2$: $5\cdot4=20$ arrangements

$4$, $3$: $5\cdot4=20$ arrangements

$4$, $2$, $2$: $5\cdot\binom42=30$ arrangements

$3$, $3$, $2$: $5\cdot\binom42=30$ arrangements

$3$, $2$, $2$, $2$: $5\cdot4=20$ arrangements

$2$, $2$, $2$, $2$, $2$: $1$ arrangement

The total is $126$, so the error seems to be in your application of the inclusion–exclusion principle.

Answer 2

It's easy to see by stars and bars that $\binom{9}{4}$ is correct. Put one ball in each box to begin with, so the question reduces to how many ways are there to put the remaining $5$ balls in $5$ boxes, and stars and bars gives $\binom{9}{4}$