Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$

Since $(x+x^2+\cdots+x^6)^{10}=x^{10}(1+x+\cdots+x^5)^{10}$ and $1+x+\cdots+x^5=\frac{1-x^6}{1-x},$ we need to find the coefficient of $x^{10}$ in $(\frac{1-x^6}{1-x})^{10}=(1-x^6)^{10}(1-x)^{-10}.$ Since $(1-x^6)^{10}(1-x)^{-10} = (1-10x^6+45x^{12}+\cdots) \sum_{m=0}^{\infty}\binom{m+9}{9}x^{m},$ the coefficient of $x^{10}$ will be $\binom{19}{9}-10\binom{13}{9}. $

The coefficient of $x^{10}$ in $(1+ x + \ldots + x^5)^{10}$ is equal to the number of integers $0 \leq x_i \leq 5 $ such that $\sum_{i=1}^{10} x_i= 10$.

We apply the Principle of Inclusion and exclusion, to deal with the restriction of $x_i \leq 5$.

If the only restriction is $0 \leq x_i$ then there are ${10 + 9 \choose 9 } $ solutions by the bars and stars method (sum of 10 non-negative integers is 10).

If $x_1 \geq 6$, then we substitute $x_1 = 6 + x_1 ^*$, and there are ${4 + 9 \choose 9}$ solutions by the stars and bars method (sum of 10 non-negative integers is 4).

Observe that we can't have 2 terms which are more than $6$.

Hence, by PIE, the coefficient is ${ 19 \choose 9} - 10 { 13 \choose 9}$, which is 85228.