The pigeonhole principle and a professor who knows $9$ jokes and tells $3$ jokes per lecture

I will assume that he doesn't tell the same joke twice (or three times) in the same lecture. Else, here is a counterexample:

Let $\{a_1, \ldots, a_9\}$ be the set of jokes. On the $i$-th day for $1 \leq i \leq 9$, tell jokes $(a_i, a_i, a_i)$. Then tell $(a_1, a_2, a_3)$, $(a_4, a_5, a_6)$, $(a_7, a_8, a_9)$ and $(a_1, a_4, a_7)$.

So, now towards the exercise. Every day the professor tells $3$ distinct pairs of jokes. Which means in total he tells $39$ pairs of jokes over the $13$ days. There are $\frac{9!}{7!2!}= 36$ pairs of jokes he can tell. So he must tell at least one pair of jokes twice.

Notice, that there are $\binom{9}{2}=36$ unique pairs of jokes.

In every lecture there are three jokes (A, B and C), thus there are three unique pairs (AB, AC, BC) per lecture used.

In series of 13 lectures there are $13*3=39$ pairs of used jokes. Thus, after the pigeonhole principle, at least one of the unique pair is used at least twice.