Four labs sharing ten hours cannot exceed a per-lab cap

Four teaching labs share ten identical instrument-hours in a week. Lab $i$ receives $x_i$ hours, where each $x_i$ is an integer satisfying $0\le x_i\le 4$, and $x_1+x_2+x_3+x_4=10$. 1. First ignore the upper bounds. Prove that the number of nonnegative integer solutions of $x_1+\cdots+x_k=n$ is $\binom{n+k-1}{k-1}$. Then, for the uniform bound $x_i\le m$, apply inclusion-exclusion: if a prescribed set of $j$ variables is at least $m+1$, the change of variables $y_i=x_i-(m+1)$ produces an unrestricted nonnegative problem. Conclude that the number of feasible tuples is \[ \sum_{j\ge 0}(-1)^j\binom{k}{j}\binom{n-j(m+1)+k-1}{k-1}, \] with the usual convention that $\binom{a}{b}=0$ when $a<b$ or $a<0$ (except $\binom{-1}{-1}$ is never needed here) and $\binom{r}{k-1}=0$ for a negative upper index in this nonnegative setting. 2. Specialize to $k=4$, $n=10$, $m=4$. Compute every surviving term explicitly and evaluate the sum. 3. The ordinary generating function for one lab is $1+x+\cdots+x^4=(1-x^5)/(1-x)$. Extract $[x^{10}](1+x+\cdots+x^4)^4$ from the expansion $(1-x^5)^4(1-x)^{-4}$ and confirm the same integer. 4. List the integer partitions of $10$ into exactly four parts, each part at most $4$ (parts may be $0$). For each type, count the distinct labellings of the four labs, and check that the type totals sum to the integer from parts 2 and 3. Explain why replacing the model by positive unknowns $y_i\ge 1$ with the same cap, or by dropping the $j=2$ term of inclusion-exclusion, produces a different number. Do not convert the problem into a marked-subset binomial-moment identity, and do not treat the variables as indistinguishable.

Without the cap, four nonnegative unknowns summing to $10$ are counted by $\binom{13}{3}$. A variable that is at least $5$ leaves a residual sum of $5$.