I was trying to count the number of natural number solutions to the equation: $x_1 + x_2 + ... + x_{11} = 20$, such that $0 \leq x_i \leq 9$, for all $i \in \{1, ..., 11\}$.
I know how to apply the first boundary condition - if I only required $0 \leq x_i$, then the number of solutions is $$\frac{30!}{20!\cdot10!}$$ because I can interpret the problem as trying to arrange $20$ balls in between $10$ bars, including the possibility of having $0$ balls in between two given bars. I just don't know how to apply the latter condition, i.e. $x_i \leq 9$.
Could anyone help me? Thanks in advance!
Big Hint:
To divide 5 into 4 parts none of which is larger than 2 (or negative), can be done in following ways (ignoring permutations)
$2,2,1,0$
$2,1,1,1$
Consider $(x^0+x^1+x^2)^4$. Then $x^5$ would be formed from terms like $x^2.x^2.x^1.x^0$ and $x^2.x^1.x^1.x^1$ only. In how many ways? Obviously that's the strength of the coefficient.