Collecting 20 dollars from different people

Question

In how many ways can I collect a total of $20$ dollars from $4$ different children and $3$ different adults, if each child can contribute up to $6$ dollars, each adult can give up to $10$ dollars, and each individual gives a non-negative whole number of dollars?

So far, I know that the generating function is $(1+x+x^2+...+x^6)^4(1+x+x^2...x^{10})^3$.

Answer 1

In Mathematica:

Coefficient[
 Sum[x^i, {i, 0, 6}]^4 Sum[x^i, {i, 0, 10}]^3, 
 x, 20]

(* 112567 *)

Answer 2

If $D_n$ is the number of ways to collect $n$ dollars from $4$ different children and $3$ different adults with each child contributing at most $6$ dollars and each adult at most $10$, then you have calculated the generating function of $D_n$: $$G(x)=\sum_{n=0}^\infty D_n x^n=(1+x+...+x^6)^4 (1+x+...+x^{10})^3$$ Since $$G(x)=D_0+D_1 x+...+D_{20}x^{20}+...+D_{54}x^{54}$$ we have that $D_{20}$, the number you are looking for, is the coefficient of $x^{20}$ in the expansion of $$(1+x+...+x^6)^4 (1+x+...+x^{10})^3$$ Can you calculate the coefficient of $x^{20}$ by hand or using a CAS?

Answer 3

There is no need to break out the calculators. Since $x+\dots+x^6=(1-x)^7/(1-x)$, and $x+\dots+x^{10}=(1-x^{11})/(1-x)$, you can write the GF as $$ (1-x)^{-7}(1-x^7)^4(1-x^{11})^3 $$ Note $$ (1-x^7)^4=(1-4x^7+6x^{14}-\dots)\\(1-x^{11})^3=(1-3x^{11}+\dots)\quad\quad\quad $$

Since we only care about the $x^{20}$ coefficient, the terms obscured by the $\dots$ can be be ignored. Therefore, the GF is $$ (1-x)^{-6}(1-4x^7+6x^{14}-\dots)(1-3x^{11}+\dots)\\=(1-x)^{-6}(1-4x^7-3x^{11}+6x^{14}+12x^{18}+\dots) $$ Finally, since $(1-x)^{-6}$ corresponds to the series $\binom{6+n-1}{6-1}$, the final anwer is $$ \binom{6+20-1}{6-1}-4\binom{6+13-1}{6-1}-3\binom{6+9-1}{6-1}+6\binom{6+6-1}{6-1}+12\binom{6+2-1}{6-1} $$