[Combinatorics]
Problem: Some students in a gym class are wearing blue jerseys, and the rest are wearing red jerseys. There are exactly 25 ways to pick a team of three players that includes at least one player wearing each color. Compute the number of students in the class.

Solution: Let $r$ and $b$ be the number of students wearing red and blue jerseys, respectively. Then either we choose two blues and one red or one blue and two reds. Thus

$$
\begin{aligned}
& \left(\begin{array}{l}
b \\
2
\end{array}\right)\left(\begin{array}{l}
r \\
1
\end{array}\right)+\left(\begin{array}{l}
b \\
1
\end{array}\right)\left(\begin{array}{l}
r \\
2
\end{array}\right)=25 \\
\Rightarrow & \frac{r b(b-1)}{2}+\frac{b r(r-1)}{2}=25 \\
\Rightarrow & r b((r-1)+(b-1))=50 \\
\Rightarrow & r b(r+b-2)=50 .
\end{aligned}
$$

Now because $r, b$, and $r+b-2$ are positive integer divisors of 50 , and $r, b \geq 2$, we have only a few possibilities to check. If $r=2$, then $b^{2}=25$, so $b=5$; the case $r=5$ is symmetric. If $r=10$, then $b(b+8)=5$, which is impossible. If $r=25$, then $b(b+23)=2$, which is also impossible. So $\{r, b\}=\{2,5\}$, and $r+b=7$.

Final answer: 7