The number of nonnegative integer solutions of $x_1+\cdots+x_6=24$ with $x_1+x_2+x_3>x_4+x_5+x_6$

Question

I try to find the number of nonnegative integer solutions of

$\begin{align} & {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}=24 \\ & {{x}_{1}}+{{x}_{2}}+{{x}_{3}}>{{x}_{4}}+{{x}_{5}}+{{x}_{6}} \\ \end{align}$

But I don't know how to refer to

${{x}_{1}}+{{x}_{2}}+{{x}_{3}}>{{x}_{4}}+{{x}_{5}}+{{x}_{6}}$

I tried to solve it : $$\begin{align} & {{x}_{1}}+{{x}_{2}}+{{x}_{3}}={{t}_{1}} \\ & {{x}_{4}}+{{x}_{5}}+{{x}_{6}}={{t}_{2}} \\ & {{t}_{1}}+{{t}_{2}}=24 \\ & 13\le {{t}_{1}}\le 24 \\ & 0\le {{t}_{2}}\le 11 \\ & ({{t}^{13}}+{{t}^{14}}+...+{{t}^{24}})(1+t+{{t}^{2}}+...{{t}^{11}})=24 \\ & 12{{t}^{24}} \\ & t{{}_{1}}>{{t}_{2}} \\ & \\ \end{align}$$

And now I am stuck.

Answer 1

An alternate method would be to use symmetry:

We have $\dbinom{29}{5}$ solutions in total, and the number of solutions where $x_1+x_2+x_3=x_4+x_5+x_6$

is given by $\dbinom{14}{2}\dbinom{14}{2}$, since then $x_1+x_2+x_3=12=x_4+x_5+x_6$.

By symmetry, half of the remaining solutions will have $x_1+x_2+x_3>x_4+x_5+x_6$,

so this gives $\displaystyle\frac{1}{2}\left[\binom{29}{5}-\binom{14}{2}\binom{14}{2}\right]=55,237$.

Answer 2

I would not bother with generating functions here. You know that $t_1$ can be any integer in $[13,24]$. For each of those values there are

$$\binom{t_1+3-1}{3-1}=\binom{t_1+2}2$$

solutions to $x_1+x_2+x_3=t_1$ in non-negative integers, and similarly

$$\binom{24-t_1+2}2=\binom{26-t_1}2$$

solutions to $x_4+x_5+x_6=24-t_1$ in non-negative integers. The total is therefore

$$\begin{align*} \sum_{t_1=13}^{24}\binom{t_1+2}2\binom{26-t_1}2&=\sum_{k=1}^{12}\binom{14+k}2\binom{14-k}2\\ &=\frac14\sum_{k=1}^{12}(14^2-k^2)(13^2-k^2)\\ &=\frac14\left(12\cdot13^2\cdot14^2-(13^2+14^2)\sum_{k=1}^{12}k^2+\sum_{k=1}^{12}k^4\right)\\ &=3\cdot13^2\cdot14^2-\frac{365}4\cdot\frac{12\cdot13\cdot25}6+\frac14\cdot\frac{12\cdot13\cdot25\cdot467}{30}\\ &=99372-\frac{118625}2+\frac{30355}2\\ &=55237\;. \end{align*}$$

There are doubtless much more elegant methods, but this required only looking up the formula for the sum of consecutive fourth powers.