Problem (Combinatorics):
Let $T=8$. At a party, everyone shakes hands with everyone else exactly once, except Ed, who leaves early. A grand total of $20 T$ handshakes take place. Compute the number of people at the party who shook hands with Ed.

Solution:
If there were $n$ people at the party, including Ed, and if Ed had not left early, there would have been $\left(\begin{array}{l}n \\ 2\end{array}\right)$ handshakes. Because Ed left early, the number of handshakes is strictly less than that, but greater than $\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$ (everyone besides Ed shook everyone else's hand). So find the least number $n$ such that $\left(\begin{array}{l}n \\ 2\end{array}\right) \geq 160$. The least such $n$ is 19 , because $\left(\begin{array}{c}18 \\ 2\end{array}\right)=153$ and $\left(\begin{array}{c}19 \\ 2\end{array}\right)=171$. Therefore there were 19 people at the party. However, $171-160=11$ handshakes never took place. Therefore the number of people who shook hands with Ed is $19-11-1=7$.

Final answer: 7