Use Inclusion-Exclusion to determine the number of integer solutions to the equation


Where $0{\leq}x_1{\leq}8; 0{\leq}x_2{\leq}5; x_3, x_4{\geq}0$.

My thought process:

I can disregard the last two restrictions because those two are given since the combination gives the number of nonnegative integer solutions.


A= # solutions given $0{\leq}x_1{\leq}8$

B= # solutions given $0{\leq}x_2{\leq}5$

Which means $A^c$ is the number of solutions for $9{\leq}x_1{\leq}14$.

*Would the possible solutions for $A^c$ then be partitioning 9+1+1+1+1+1=14? Or rather, give us ${6+4-1\choose 4-1}$?

*And $B^c$ is the number of solutions for $6{\leq}x_2{\leq}14$, so we would have $6+1+1+1+1+1+1+1+1=14$ or ${9+4-1\choose 4-1}$?

If $x_1{\gt8}$ and $x_2{\gt}5$ then for $A^c{\cap}B^c$ we have $$9+6+0+0=15$$ or similar and we see there are no possible solutions for $x_1+x_2+x_3+x_4=14$.

Then $$A{\cap}B = A{\cup}B-A^c-B^c+A^c\cap B^c$$ $$={14+4-1\choose 4-1} - {6+4-1\choose 4-1}- {9+4-1\choose 4-1} +0$$ $$=376$$

*I'm a little nervous about my reasoning for $A^c$ and $B^c$. I always end up missing some crucial step in these kinds of problems. - resolved in comments


You have the right idea, but you’re a little off. The solutions that have $x_1\ge 9$ correspond to the non-negative solutions of $y+x_2+x_3+x_4=5$, where $y=x_1-9$, so there are $\binom{5+4-1}{4-1}$ of them, not $\binom{6+4-1}{4-1}$. Similarly, there are $\binom{8+4-1}{4-1}$ solutions with $x_2\ge 6$. The rest is fine.

  • $\begingroup$ Thank you!! I have a hard time wrapping my head around counting concepts. $\endgroup$ – Karen M. Feb 23 '15 at 22:19
  • $\begingroup$ @Karen: You’re welcome! You don’t seem to be doing too badly with them, judging by the questions that you’ve asked. $\endgroup$ – Brian M. Scott Feb 23 '15 at 22:21

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