Number of solutions in non-negative integers to $x_1 + x_2 + x_3 + x_4 + x_5 = 9$ where $x_1, x_2,x_3,x_4,x_5 \neq 1$ using generating functions

Question

Using generating functions, the answer is the coefficient of $x^9$ in the expression $(1+x^2 +x^3 +x^4 \dots)^5$

Since $1+x+x^2+x^3 + \dots = \frac{1}{1-x}$ we get:

$(1+x^2 +x^3 +x^4 \dots)^5 = (\frac{1}{1-x} - x)^5 = (\frac{1 - x + x^2}{1-x})^5 = (\frac{1}{1-x})^5 \cdot (1+x^2-x)^5$

Since we generally have $(\frac{1}{1-x})^n = \sum_{k=0}^n {n+k-1 \choose k}x^k$

We get $(\frac{1}{1-x})^5 = \sum_{k=0}^5 {5+k-1 \choose k}x^k = {4 \choose 0} + {5 \choose 1}x+{6 \choose 2}x^2+{7 \choose 3}x^3+{8 \choose 4}x^4+{9 \choose 5}x^5$

And by the binomial theorem we get $(1+x^2-x)^5 = \sum_{k=0}^5 {5 \choose k}(x^2-x)^k = {5 \choose 0} + {5 \choose 1}(x^2-x) + {5 \choose 2}(x^2-x)^2 + {5 \choose 3}(x^2-x)^3 + {5 \choose 4}(x^2-x)^4 + {5 \choose 5}(x^2-x)^5$

So we are left with calculating the coefficient of $x^9$ in

$({4 \choose 0} + {5 \choose 1}x+{6 \choose 2}x^2+{7 \choose 3}x^3+{8 \choose 4}x^4+{9 \choose 5}x^5) \cdot ({5 \choose 0} + {5 \choose 1}(x^2-x) + {5 \choose 2}(x^2-x)^2 + {5 \choose 3}(x^2-x)^3 + {5 \choose 4}(x^2-x)^4 + {5 \choose 5}(x^2-x)^5)$

The problem is that the $(x^2-x)^k$ parts are really long to compute and it seems like there should be an easier way. Any advice on how to solve this without having to open up these expressions?

Answer 1

Hint: we have to find the coefficient of $x^9$ in $${(1-x)}^{-5}{(x^2-x+1)}^5$$ which can be transformed to by multiplying denominator and numerator by ${(1+x)}^5$ $${(x^3+1)}^5{(1-x^2)}^{-5}$$ or $${(1+5x^3+10x^6+10x^9..)}{(1+5x^2+15x^4+35x^6...)}$$