number of positive integer solutions of $x+y+z+w=24$

Question

Determine the number of positive interger solutions of $x+y+z+w=24$ such that $x\leq 6, y \leq 7, z\leq 8, w\leq 9$ My try: I used generating polynomial as $$f(x)=(x+x^2+\cdots+x^6)(x+x^2+\cdots+x^7)(x+x^2+\cdots+x^8)(x+x^2+\cdots+x^9$$ $\implies$ $$f(x)=x^4(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}$$ We need to collect the coefficient of $x^{20}$ in $(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}$ So we can write $$(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}=(1-x)^{-4}(1-(x^6+x^7+x^8+x^9)+(x^{13}+x^{14}+2x^{15}+x^{16}+x^{17})-..)$$ Also using $$(1-x)^{-4}=\sum_{k=0}^{\infty}\binom{k+3}{k}x^k$$ We get required coefficient of $x^{20}$ as $$\binom{23}{3}-\left(\binom{14}{11}+\binom{15}{12}+\binom{16}{13}+\binom{17}{14}\right)+\left(\binom{6}{3}+\binom{7}{3}+2\binom{8}{3}+\binom{9}{3}+\binom{10}{3}\right)=83$$ But the answer is not matching.

Answer 1

Your answer is right here is an easier way to go about this which may bring less calculation errors

set $x=6-u.y=7-v,z=8-t,w=9-p$

then with $0\le u\le 5,0\le v\le 6,0\le t\le 7,0\le p\le 8 $ we have to find the number of solutions of $$u+v+t+p=6$$ but $u\neq 6$ hence number of solutions is $\binom{6+4-1}{4-1}-1=83$