Number of solutions

Let $x_i = -3 + y_i, y_i \geq 0$ for $i = 1, \ldots, 4$ . Substituting into the original equation yields $y_1 + y_2 + y_3 + y_4 = 13$ The number of non-negative integral solutions to the equation above is $\binom{13 + 3}{3} = 560$ . Since the problem requires that $y_i \leq 6$ , we have to find the number of solutions violating that requirement. Notice that only one variable at a time can violate the requirement.

So let us consider the case where one of the variables, say $y_1$ is greater than 6. We express this by letting $y_1 = 7 + z_1$ where $z_1 \geq 0$ , and get $z_1 + y_2 + y_3 + y_4 = 6$ The number of non-negative solutions to the equation above is $\binom{6 + 3}{3} = 84$ . We apply the same argument to $y_2, y_3$ and $y_4$ , respectively.

So the number of solutions to $x_1 + x_2 + x_3 + x_4 = 1, \quad x_i \in \{-3, \ldots, 3\}$ is $\binom{16}{3} - 4\cdot \binom{9}{3} = \boxed{224}$

A slight alteration to this script would provide a list of the solutions too.

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from itertools import product

poss=range(-3,4)

count=0
for x in product(poss,poss,poss,poss):
    count+=sum(x)==1

print count