On some exam i had task "Show that number of partitions of $n$ on four parts is equal to number of partitions of number 3n on four parts, but each part not greater than $n-1$"
So first is $$n = a + b + c + d$$ where $a,b,c,d>0$
Second $$3n=a+b+c+d$$ where $0<a,b,c,d<n$
So i thought i might represent it with generating functions for first answer will be $$[x^n] (x+x^2+x^3+\dots)^4=$$ $$[x^n]x^4 (1+x+x^2+x^3+\dots)^4=$$ $$[x^{n-4}]\frac{1}{(1-x)^4}=\binom{n-1}{3}$$
And same thing with second equation. $$[x^{3n}](x+x^2+x^3+\dots+x^n)^4=$$ $$[x^{3n}]x^4(1+x+x^2+\dots+x^{n-1})^4=$$ and so on
Why is this solution wrong?? I was sure that it's good solution, but for some reason i got negative response. I'd love to get some answer from you guys before i eventually make a complainment, because i don't want to complain if my thinking is totally wrong. Thanks in advance!
You could write the second problem in the equivalent form $ n= (n-a)+(n-b)+(n-c)+(n-d)$ and observe that it is, in this way, equivalent to the first one.
Edit: Assuming that the computation that follows after (or instead of) "and so on" is done correctly, I believe that your approach is correct [I've done the computation myself and I've got the same answer as you].
Edit 2: As Michael mentioned, the last power of the $3n$ case is $n-1$, not $n$. My computation was with $n-1$, sorry for not paying attention to this detail. Otherwise, the method is sound.