1. **Problem statement:** We have 400 students with scores from 0 to 10. For part (a), we know for each $i \in \{1,2,3,4,5\}$, the number of students scoring $i+5$ is 2 more than those scoring $i$. We want to find how many different histograms (distributions of students per score) satisfy this. 2. **Define variables:** Let $x_i$ be the number of students scoring $i$ for $i=0,1,\ldots,10$. 3. **Given conditions for (a):** For $i=1$ to $5$, $$x_{i+5} = x_i + 2$$ 4. **Sum of all students:** $$\sum_{i=0}^{10} x_i = 400$$ 5. **Express $x_6$ to $x_{10}$ in terms of $x_1$ to $x_5$:** $$x_6 = x_1 + 2,\quad x_7 = x_2 + 2,\quad x_8 = x_3 + 2,\quad x_9 = x_4 + 2,\quad x_{10} = x_5 + 2$$ 6. **Rewrite sum:** $$x_0 + \sum_{i=1}^5 x_i + \sum_{i=6}^{10} x_i = x_0 + \sum_{i=1}^5 x_i + \sum_{i=1}^5 (x_i + 2) = x_0 + 2\sum_{i=1}^5 x_i + 10 = 400$$ 7. **Simplify:** $$x_0 + 2\sum_{i=1}^5 x_i = 390$$ 8. **Non-negativity constraints:** All $x_i \geq 0$ and integers. 9. **Count solutions:** Let $S = \sum_{i=1}^5 x_i$. Then $$x_0 + 2S = 390$$ Since $x_0 \geq 0$, $2S \leq 390 \Rightarrow S \leq 195$. 10. For each fixed $S$, number of solutions to $x_0 = 390 - 2S$ is 1 (since $x_0$ is determined). 11. Number of integer solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = S$ with $x_i \geq 0$ is $$\binom{S+5-1}{5-1} = \binom{S+4}{4}$$ 12. Total number of histograms is $$\sum_{S=0}^{195} \binom{S+4}{4}$$ 13. Use the identity: $$\sum_{k=0}^n \binom{k+r}{r} = \binom{n+r+1}{r+1}$$ Here $r=4$, so $$\sum_{S=0}^{195} \binom{S+4}{4} = \binom{195+4+1}{5} = \binom{200}{5}$$ 14. **Final answer for (a):** $$\boxed{\binom{200}{5}}$$ --- **Slug:** histogram conditions **Subject:** combinatorics **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 6