Using generating functions only:
In how many ways we can 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?
Answer:
The generating function is
$ F(x)=(1+x+x^2+......+x^6)^4 \cdot (1+x+x^2+.......+x^{10})^3 \ $
We have to find the coefficient of $ \ x^{20} \ $ from the expansion of the generating function.
But I am unable to find the coefficient of $ \ x^{20} \ $.
continuing and applying geometric series summation$$F(x)=\left(\frac{x^7-1}{x-1}\right)^4\left(\frac{x^{11}-1}{x-1}\right)^3=(x^7-1)^4(x^{11}-1)^3(x-1)^{-7}$$
$$F(x)=\left(\sum\binom4kx^{7k}(-1)^{4-k}\right)\left(\sum\binom3ix^{11i}(-1)^{3-i}\right)\left(\sum\binom{-7}jx^j(-1)^{-7-j}\right)$$ $$F(x)=\left(\sum\binom4kx^{7k}(-1)^{k}\right)\left(\sum\binom3ix^{11i}(-1)^{i+1}\right)\left(\sum\binom{7+j-1}6x^j(-1)^{-7-j+j}\right)$$ $$F(x)=-\left(\sum\binom4kx^{7k}(-1)^{k}\right)\left(\sum\binom3ix^{11i}(-1)^{i+1}\right)\left(\sum\binom{6+j}6x^j\right)$$
now find all possible solution of $7k+11i+j=20$ with condition $0\le k\le4,0\le i\le3,0\le j$
Negative Binomial Theorem