[Combinatorics]
Problem: Let $T=75$. At Wash College of Higher Education (Wash Ed.), the entering class has $n$ students. Each day, two of these students are selected to oil the slide rules. If the entering class had two more students, there would be $T$ more ways of selecting the two slide rule oilers. Compute $n$.

Solution: With $n$ students, Wash Ed. can choose slide-rule oilers in $\left(\begin{array}{l}n \\ 2\end{array}\right)=\frac{n(n-1)}{2}$ ways. With $n+2$ students, there would be $\left(\begin{array}{c}n+2 \\ 2\end{array}\right)=\frac{(n+2)(n+1)}{2}$ ways of choosing the oilers. The difference is $\frac{(n+2)(n+1)}{2}-\frac{n(n-1)}{2}=T$. Simplifying yields $\frac{\left(n^{2}+3 n+2\right)-\left(n^{2}-n\right)}{2}=2 n+1=T$, so $n=\frac{T-1}{2}$. Because $T=75, n=37$.

Final answer: 37