ordered partitions with restrictions

Question

Can these two questions be solved only with a computer? (1)Using ten numbers ranging from 0 to 10 and allowed to appear more than once, how many ordered partitions are there so that the sum equals 53? (2)Using eight numbers ranging from 4 to 7 and allowed to appear more than once, how many ordered partitions are there so that the sum equals 37?

Answer 1

Assuming that my interpretation of ordered partition is the correct one, you are simply looking for (1) the coefficient of $x^{53}$ in $(1+x+x^2+\ldots+x^{10})^{10}$ and (2) the coefficient of $x^{37}$ in $(x^4+x^5+x^6+x^7)^8$. About (1): $$\begin{eqnarray*} [x^{53}](1+x+x^2+\ldots+x^{10})^{10}&=&[x^{53}]\frac{(1-x^{11})^{10}}{(1-x)^{10}}\\&=&[x^{53}]\sum_{k=0}^{10}\binom{10}{k}(-1)^k x^{11k}\sum_{j\geq 0}\binom{9+j}{9}x^j\\&=&\sum_{k=0}^{4}(-1)^k\binom{10}{k}\binom{62-11k}{9}\\&=&\color{red}{976626970}\end{eqnarray*}$$ and about (2): $$\begin{eqnarray*}[x^{37}](x^4+x^5+x^6+x^7)^8 &=& [x^5](1+x+x^2+x^3)^8\\&=&[x^5]\frac{(1-x^4)^8}{(1-x)^8}\\&=&[x^5]\sum_{k=0}^{8}\binom{8}{k}(-1)^k x^{4k}\sum_{j\geq 0}\binom{7+j}{7}x^j\\&=&\sum_{k=0}^{1}(-1)^k\binom{8}{k}\binom{12-4k}{7}\\&=&\color{red}{728}.\end{eqnarray*}$$

If ordered partition just means partition, the answers to (1) and (2) are given by

$$ [y^{10}x^{53}]\prod_{k=1}^{10}\frac{1}{1-y x^{k}}=\color{red}{2897},\qquad [y^8 x^{13}]\prod_{k=1}^{4}\frac{1}{1-y x^k}=\color{red}{5}. $$