TITLE: The pigeonhole principle and a professor who knows $9$ jokes and tells $3$ jokes per lecture
QUESTION [41 upvotes]: A professor knows $9$ jokes and tells $3$ jokes per lecture.
  Prove that in a course of $13$ lectures there is going to be a pair of jokes that will be told together in at least $2$ lectures.

I've started with counting how many possibilities there are to tell jokes in a lecture. Let
$$J := \{1,2,\dots,9\}$$
The amount of all different possible combinations for jokes is $9 \choose 3$ and for each lecture there are going to be $3$ unique pairs of jokes $\left(\frac{3!}{2!}=3\right)$.
I'm not sure how to continue from here to get to the PHP, I think I might be doing something wrong here, any advice how to abstract it properly?

This is an exercise from the Tel-Aviv University entry test preparation and I'm not a student yet so elementry combinatorics should do here.

REPLY [4 votes]: (Please forgive my inexperience with using MathJax)
It should be assumed that the professor never tells the same joke twice in a single lecture.  I think this is reasonable, and it seems that other respondents are also assuming this.  If a joke can be repeated during a lecture, then I think it adds a layer of complexity to the problem that is uncharacteristic of traditional pigeonhole principle problems.  
Amount of possible joke-pairs = ${9 \choose 2}$ = 72.
Amount of joke-pairs told during a lecture = ${3 \choose 2}$ = 3.  This is the first-third joke pairing, the first-second joke pairing, and the second-third joke pairing.
Total amount of joke-pairs that occur = 13 lectures * 3 per lecture = 39.  
According to the pigeonhole principle. If 72 > 39, then at least one joke-pair will be repeated.