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In a party people shake hands with one another (not necessarily everyone with everyone else).
(a) Show that 2 people shake hands the same no. of times.
(b) Show that the number of people who shake hands an odd no. of time is even.

I find this question to be very tough and I can't find a way to even start answering this. It would be great if I could get a small hint in order for me to at least start it.

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    @TakahiroWaki This is a common exercise in combinatorics and pigeon-hole principle. The OP may not have included a key phrase that noone shakes hands with someone else more than once each. This allows us to describe the scenario as a simple graph (not a multigraph), and the result can be rephrased "any degree sequence in a graph will have a repeated entry" There is no counterexample because it is true. – 2017-01-16
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    I found that to shake hands with same person is at most only once. It seem that I have counted multiple times. – 2017-01-16

3 Answers 3

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Hints:

For (A) with $n$ people at the party, what are the possible number of hands each person could shake?

Show that it is impossible for both someone to have shaken hands with everyone and someone to have shaken hands with noone simultaneously.

Is there some way to use the pigeon-hole principle here?

For (B) count the number of handshakes that occur by adding up how many handshakes people participated in individually and recognize that this overcounted somehow (because each handshake we counted twice: once from the shorter person of the two and again from the taller person). This gives us a useful result called "the handshaking lemma."

What happens then if there are an odd number of people who shook an odd number of hands? Remember that the number of handshakes must be a whole number.

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    Got part B but could you elaborate a bit on A – 2017-01-16
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    @OsheenSachdev Have you ever heard of the [pigeon-hole principle](https://en.wikipedia.org/wiki/Pigeonhole_principle)? If you have only $k$ holes and strictly more than $k$ pigeons and place each pigeon in a hole, at least one hole must have strictly more than one pigeon in it. How many possibilities are there for numbers of handshakes for an individual to participate in? How many people are there shaking hands? – 2017-01-16
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    Still didn't get it :( – 2017-01-16
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    @OsheenSachdev if there are $n$ people, each person could shake hands with $0$ people, $1$ person, $2$ people,... on up to shaking hands with $n-1$ people. Count how many different answers there are to asking the person the question "How many hands did you shake?" How many people are there? If the people are the pigeons, and the possible answers to the question "how many hands did you shake" are the holes, can we conclude anything yet? No? How about now noticing that at least one of the holes "I shook hands with noone" or "I shook hands with everyone" has to be empty... now what? – 2017-01-16
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    Oh no, I got the fact that both statements can't be true simultaneously but how do we reach the conclusion that 2 people shake hands the same no. of times? – 2017-01-16
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    "*Since there are more pigeons than holes there must be a hole with at least two pigeons in the same hole*" Now, replace the word "pigeons" and "holes" with the appropriate terms for the context of your specific question, remember we are talking about people and number of handshakes they participated in – 2017-01-16
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    Oh! Got it now, Thank you! – 2017-01-16
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$(a)$ Say that there is $n$ people in the party and assume everyone shakes hands at least once. The number of handshakes possible for everybody then is just $n-1$ (self-handshaking doesn't count) while there is $n$ people in the party, so the pingeon-hole principle gives the desired result. Now, if we don't assume that everyone shakes at least one hand then we can separate the $n$ people between $k$ which don't shake any hands and $m$ which do ($k+m=n$), and the problem is reduced to the last case (with $n=m$).

$(b)$ If a person $A$ shakes hands with a person $B$, then $B$ shakes hands with $A$. Silly as it sounds this tells us that we can count handshakes in pairs, so the total number of handshakes must be even. Counting separately the handshakes of those who have given an even number of handshakes and those who have given an odd number of handshakes leads to the desired conclusion.

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    If the number of all handshakes is n(n-1), this number is even in any case because the product of an even number with an odd number is always even. If n is even, n-1 is odd and if n is odd, n-1 is even. – 2017-01-16
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    @Uwe The total number of handshakes might not be $n(n-1)$. – 2017-01-16
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For (B) try induction. What happens when the next two people shake hands?