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.

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6 Answers6


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.

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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.

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Jaroslaw Matlak
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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 $8$.

Carsten S
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  • Could you clarify where these numbers are coming from? – Ray Jun 07 '18 at 16:17
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    @Ray, I hope that it is better now. – Carsten S Jun 07 '18 at 19:49
  • I really like reasoning in this answer :) – Jaroslaw Matlak Jun 08 '18 at 06:10
  • +1 for the alternate viewpoint, but I think this answer would benefit from an explanation of how the reasoning fails if we don't assume that the professor is telling three *distinct* jokes per lecture. (In that case, of course, we need not "have (at least) $5\cdot2$ slots to fill". I say "of course", but I'm not convinced that I would have noticed had I not read the accepted answer, and I believe that answers should be able to stand on their own.) – A. Howells Jun 08 '18 at 13:57

Let $a_1,a_2,a_3$ be the jokes he told on first day. To tell jokes for 13 days,without running out of jokes he has to tell jokes efficiently.

So he takes one joke from first day and repeats it. Second day jokes are {$a_1,a_4,a_5$ } ...like that maximum number of days he can do that is=(9-1)/2=4 days

Now he fixes $a_2$,repeat the above process without including $a_1,a_3$. Maximum number of days he can do that is=(9-3)/2=3 days.

Repeating this,we get maximum number days he can afford to tell joke=4+3+2+1=10 days

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    You've shown that one particular joke schedule results in repetitions but you haven't shown that this schedule delays repetitions for as long as possible. – David Richerby Jun 08 '18 at 14:26

(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.

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    Your first result is the number of permutations, rather than combinations, possible. If you switch it to combinations, you will get 36. Additionally: your equality is reversed: 10 possible dog breeds and I choose 5 does not pidgenhole me into choosing two of the same breed. Possible must be less than total choices to guarantee a repeat. – TemporalWolf Jun 08 '18 at 21:44

In each one of the $13$ lectures, the professor can choose from a total of ${9 \choose 3 } = 84$ different combinations of jokes.

And, in each particular one of these $84$ combinations, there are ${ 3 \choose 2 } = 3$ distinct pairs.

Thus in each one of the $13$ lectures, there are $84 \times 3 = 252$ different pairs of jokes possible.

On the other hand, there are only ${ 9 \choose 2} = 36$ different pairs of jokes that could possibly be chosen from a total of $9$ jokes.

Hence we must have repetitions.

Hope this helps.

Saaqib Mahmood
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    Notice, that in these 84 combinations the pairs of jokes could repeat (ABC, ABD, ACD,..). Thus in every lecture there are only 36 possible unique pairs of jokes, not 252. – Jaroslaw Matlak Jun 07 '18 at 11:34
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    @JaroslawMatlak you're absolutely right. I totally agree. Thank you. – Saaqib Mahmood Jun 07 '18 at 12:04
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    I don't understand. You compute what should be the same number ("different pairs of jokes possible") in two different ways, and get two different answers. Doesn't that just mean that your working was wrong? – David Richerby Jun 08 '18 at 14:23
  • @DavidRicherby there are two possibilities: either my working is wrong, as you've suggested, or the two numbers cannot be equal. You see, two different combinations of $3$ jokes out of $9$ can include the same pair of jokes, thus the counting in the same pairs of jokes multiple times. – Saaqib Mahmood Jun 09 '18 at 02:26
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    You never actually use the fact that there are thirteen lectures so, if valid, your proof would imply that there must be repetitions in a course of any number of lectures, including one. Clearly, there does _not_ have to be repetition in a single lecture, so your proof is invalid. Allow me to be a little less delicate than in my first comment. You calculated the same number twice and got two different answers. The first one was wrong, as Jaroslaw has already told you, which you “totally agree” with. If you know your answer is wrong, you should delete it. – David Richerby Jun 09 '18 at 08:45