Some possible interpretations I can see:
- How many integer solutions exist to the system $x_1+x_2+x_3+x_4+x_5=31$ such that $x_i\leq i$ for all $i\in \{1,2,\dots,5\}$
There are no solutions since:
$$x_1+x_2+x_3+x_4+x_5\leq 1+2+3+4+5=15<31$$
seen by observing the upper bounds on each of $x_i$ individually.
- How many integer solutions exist to the system $x_1+x_2+x_3+x_4+x_5=31$ exist such that $x_i\leq i$ for some $i\in\{1,2,\dots,5\}$
Notice that $x_1=31, x_2=x_3=x_4=x_5=0$ is a solution and also $x_1=31, x_2=k, x_3=-k, x_4=x_5=0$ is a solution for all values of $k$. This also satisfies the condition that at least one of the $x_i$ is at most $i$ since $x_5=0<5$ in all cases
- How many non-negative integer solutions exist to the system $x_1+x_2+x_3+x_4+x_5=31$ such that at least one of the $x_i$ is bounded above by $i$
Approach via inclusion-exclusion. Letting $A_i$ be the event that $x_i>i$ we are tasked with finding $|U\setminus(A_1^c\cup\dots \cup A_5^c)|=|U|-|A_1\cap\dots \cap A_5|$
To find $|A_1\cap\dots\cap A_5|$, we have the system that: $x_1+\dots+x_5=31$ and $i<x_i$ implying $i+1\leq x_i$. Via a change of variable, setting $y_i = x_i-i-1$, we have $y_1+\dots+y_5=11$. There are then $\left(\!\!\binom{11}{5}\!\!\right)=\binom{11+5-1}{5-1}=\binom{15}{4}$ possible solutions.
Going back to the original problem, there are then $\binom{31+5-1}{5-1}-\binom{11+5-1}{5-1}=\binom{35}{4}-\binom{15}{4}$ total number of solutions.
- How many integer solutions exist to the system $x_1+x_2+x_3+x_4+x_5=31$ such that $x_i\geq i$ for all $i$
Again, approach via a change of variable by setting $y_i=x_i-i$. You have then $\binom{16+5-1}{5-1}=\binom{20}{4}$ total solutions.