Find number of ways to split $1$ dollar into $5$, $10$, $20$, $50$ cents
I am going to use generating functions: $$n = [x^{100}] (1+x^5+x^{10}+\cdots)(1+x^{10}+x^{20}+\cdots)(1+x^{20}+x^{40}+\cdots)(1+x^{50}+x^{100}+\cdots) = \\ [x^{100}]\frac{1}{1-x^5}\frac{1}{1-x^{10}}\frac{1}{1-x^{20}}\frac{1}{1-x^{50}} $$ Ok, I know that computers are able to solve that but how can I easily get coefficient at $x^{100}$?
Let $$\begin{align}f(y)&=\frac{1}{1-y}\frac{1}{1-y^2}\frac{1}{1-y^4}\\&=\frac{(1+y+y^2+y^3)(1+y^2)}{(1-y^4)^3}\\&=\frac{1+y+2y^2+2y^3+y^4+y^5}{(1-y^4)^3}\end{align}$$
Letting $p(y)=1+y+2y^2+2y^3+y^4+y^5$, then $$[y^{20}]f(y)\frac{1}{1-y^{10}} = [y^0]f(y)+[y^{10}]f(y)+[y^{20}]f(y)$$
Now use that $$\frac{1}{(1-y^4)^3}=\sum_{j=0}^{\infty} \binom{j+2}{2}y^{4j}$$
So $$\begin{align}[y^0]f(y)&=1\\ [y^{10}]f(y)&=\binom{0+2}{2}[y^{10}]p(y) + \binom{1+2}{2}[y^{6}]p(y)+\binom{2+2}{2}[y^2]p(y)\\&=0+0+6\cdot2\\&=12\\ [y^{20}]f(y)&=\binom{4+2}{2}[y^4]p(y)+\binom{5+2}{2}[y^0]p(y)\\ &=15\cdot 1 + 21\cdot 1\\ &=36\end{align}$$
So your result is $1+12+36=49.$
You have in general that $$\begin{align}[y^{4j}]f(y)&=\binom{j+1}{2}+\binom{j+2}{2}=(j+1)^2\\ [y^{4j+1}]f(y)&=\binom{j+1}{2}+\binom{j+2}{2}=(j+1)^2\\ [y^{4j+2}]f(y)&=2\binom{j+2}{2}=(j+2)(j+1)\\ [y^{4j+3}]f(y)&=2\binom{j+2}{2}=(j+2)(j+1) \end{align}$$