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.

Sometimes it is a bit easier to prove a bit more: We can actually show that the joke that has been told most often must be in the pair of jokes that we are looking for.

There will be at least one joke that has been told at least 5 times: There are $13\cdot3$ slots for jokes and if each joke was told at most 4 times, 9 jokes could only fill $9\cdot 4$ slots, but $13\cdot 3 > 9\cdot 4$. (Indeed I calculated: With 12 days each joke would be told an average $\frac{12\cdot3}9=\frac{12}3=4$ times, so with one day more at least one joke has to be told more than 4 times.)

Now if we restrict our attention to only the days on which that joke has been told, then of the other 8 jokes, at least one has been told twice: We have (at least) $5\cdot2$ slots to fill, and that is more than $8$.