For example I have to break $\$10,000$ into denominations of $\$1, \$5, \$10, \$100, \$500, \$1000$ with not using only total of $500$ denominations and each denomination not more than $9$.
Problem 1) Face value: Distribute $500$ into $6$ places with non-negative value. Coefficient of $500$ in
$f(z) = (1 + x + x^2 + x^3 + ... + x^9)^6$.
Problem 2) Place value: Total place value should be $10000$, coefficient of $10000$ in
$f(z) = (1 + x + x^2 +x^3 + ... + x^9)(1 + x^5 + x^{10} + x^{15} + ... + x^{45})(1 + x^{10} + x^{20} + ... + x^{90})(1 + x^{100}+ ... + x^{900})(1 + x^{500} + x^{1000} + ... + x^{4500})(1 + x^{1000} + ... + x^{9000})$
How do I combine these two functions?
Clearer version suggested by BMS:
We have a supply of bills in the six denominations $\$1,\$5,\$10,\$100,\$500$, and $\$1000$, and we wish to count the ways to form a total of $\$10000$. We must use exactly $500$ bills, and we may not use any denomination more than $9$ times. If we didn’t care about the total value of the $500$ bills, the answer would be the coefficient of $x^{500}$ in $(1+x+x^2+\ldots+x^9)^6$. If we didn’t care about the number of bills used, either in total or by denomination, the answer would be the coefficient of $x^{10000}$ in the product of the six polynomials $1+x^d+x^{2d}+\ldots+x^{9d}$ as $d$ runs over the set $\{1,5,10,100,500,1000\}$.
How can I combine these approaches to solve the actual problem, taking into account all of the restrictions?