How many integer solutions does the following equation have? $x_1 + x_2 + x_3 + x_4 < 50$

Question

Question:
Compute how many different integer solutions the following inequality has: $$ x_1 + x_2 + x_3 + x_4 < 50 $$
when we require that $-1 \leq x_1, -2 \leq x_2 \leq 2, -3 < x_3, -4 < x_4 < 4.$

My approach:
So I will start by making the inequality to equality. $$ x_1 + x_2 + x_3 + x_4 < 50 $$ $$ x_0 + x_1 + x_2 + x_3 + x_4 = 49 $$ $$x_0 \geq 0 $$ Then I would change $x_i$ for $y_i$:
$$y_1 - 1 = x_1$$ $$y_2 - 2 = x_2$$ $$y_3 - 2 = x_3$$ $$y_4 - 3 = x_4$$ So that would be: $$y_0 + y_1 - 1 + y_2 - 2 + y_3 - 2 + y_4 - 3 = 49 $$ $$y_0 + y_1 + y_2 + y_3 + y_4 = 57 $$ Since I have bounds on $x_2$ and $x_4$, I need to make condition for them: $$-2 \leq x_2 \leq 2$$ $$0 \leq x_2 \leq 4$$ $$x_2 \geq 5$$ $$and$$ $$-4 < x_4 < 4$$ $$0 < x_4 < 8$$ $$x_4 > 8$$ And finally to count it: $$\binom{57+5-1}{5-1}...$$
But I don't know how to get the final number (including those conditions). Is it possible to do it with inclusion–exclusion principle ?
Any help on what to do would be appreciated.

Answer 1

You're on the right track.

Hint We're aiming to solve the inequality $$\phantom{(\ast)} \qquad y_1 + y_2 + y_3 + y_4 \leq N,$$ or just as well the equality $$\phantom{(\ast)} \qquad y_0 + y_1 + y_2 + y_3 + y_4 = N , \qquad (\ast)$$ in nonnegative integers.

As you suggest, we can use the Inclusion-Exclusion Principle: Let $S$ denote the set of solutions $(y_0, \ldots, y_4)$ of $(\ast)$, $S_2 \subset S$ the set of solutions satisfying $y_2 \geq a$, and $S_4 \subset S$ the set of solutions satisfying $y_4 \geq b$, so that our aim is to compute $|S| - |S_2 \cup S_4| = |S| - |S_2| - |S_4| + |S_2 \cap S_4|$.

The number of solutions of $(\ast)$ is $$|S| = {{N + 4} \choose 4}$$ (see this question for a derivation and OEIS 000332 for the sequence).

We now consider the number of solutions of $(\ast)$ in $S_2$, that is, that satisfy $y_2 \geq a$ (in particular not imposing the condition on $y_4$). If we set $\color{#ff0000}{z_2} := y_2 - a$, then we are just as well looking for the number of solutions in nonnegative integers to $$y_0 + y_1 + \color{#ff0000}{z_2} + y_3 + y_4 = N - a ,$$ which we know from the computation of $|S|$ is $$|S_2| = {{N - a + 4} \choose 4} .$$

By symmetry of notation, $$|S_4| = {{N - b + 4} \choose 4} .$$ For essentially the same reason, $$|S_2 \cap S_4| = {{N - (a + b) + 4} \choose 4} .$$ In the end, we find that the number of solutions to $(\ast)$ in nonnegative integers subject to our constraints on $y_2$, $y_4$ are $$|S| - |S_2 \cup S_4| = {N + 4 \choose 4} - {{N - a + 4} \choose 4} - {{N - b + 4} \choose 4} + {{N - (a + b) + 4} \choose 4} .$$ In our particular case, $N = 57$, $a = 5, b = 7$, giving $${61 \choose 4} - {56 \choose 4} - {54 \choose 4} + {49 \choose 4} = \boxed{50\,190} ,$$ which agrees the value produced by the script in Marco's answer.

Answer 2

You can simply brute force it.

For the bounds on $x_2,x_3,x_4$ we have to have $x_1<57$.

For the bounds on $x_1,x_2,x_4$ we have to have $x_3<56$

Now that all the numbers are on a finite interval of integers I have written a little Python script that check all the possible sums. The final result should be $$ 50190 $$