# Combinatorics question: Prove 2 people at a party know the same amount of people

I recently had an assignment and got this question wrong, was wondering what I left out.

Prove that at a party where there are at least two people, there are two people who know the same number of other people there. Be sure to use the variable "n" when writing your answer.

n >= 2 Case1: another person comes to the party, but doesn't know either of the first two. So the original two still only know the same number of people. Case2: another person comes, and knows one out of the original 2, so therefore the >newcommer, and the one that doesnt know the newcommer both know the same number of people. Case 3: another person comes and knows both of them, implying that they all know each >other, and therefore they all know the same number of people.

So therefore if n>=2, where n and n-1 know each other, in either case whether n+1 joins, >there will be at least two people who know the same amount of people.

Many thanks in advance. Have a test coming up.

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I understand case 1 and case 3, but how do you get "the one that doesn't know and the newcommer both know the same number of people" in case 2? –  anon Aug 1 '12 at 3:54
Ah I see what you are saying. So I guess if the original people at the party (A and B) know each other, and C knows either A or B, then C knows 1 other person, and the one C didn't chose, knows that same person. And if C joins, and A or B do not know each other, then C will know either A or B, and A or B will know C –  treehau5 Aug 1 '12 at 3:56
Are you talking about general $n$ or just $n=2$? You realize you have to prove it for all $n$, not just a few cases like $2$ and $3$, right? –  anon Aug 1 '12 at 4:04
Suppose A,B, and C are already there, with A and B knowing each other. So we have A and B knowing 1 and C knowing 0. If D arrives, knowing only A, now B and D are the pair who know the same number. I don't see how to make induction show this-it seems "too random". –  Ross Millikan Aug 1 '12 at 4:04

You are trying to prove this by induction, but it is not easy to do it that way. Say you had a party where two people each know four others. Now if a new guest arrives who knows one of them and not the other, you have broken the pair. You have to show that there is a pair in the new party.

I would approach it directly. If there are $n$ people at the party, each one knows somewhere from $0$ to $n-1$ of the others, which is $n$ possibilities. But if one person knows all the others, nobody knows nobody else (we are assuming knowing is reflexive here, otherwise the statement is not true-each person could know precisely those who arrived before him). So the number each person knows is either in the range $[0,n-2]$ or $[1,n-1]$. In either case, the pigeonhole principle says there are two that match.

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Good answer, but the OP asked for a solution along his lines.. –  Rijul Saini Aug 1 '12 at 3:59
Ah okay, that was the kicker. I was saying n was the number of people at the party, you are saying n is the number of possibilities. When I take that approach it makes much more sense, thank you. –  treehau5 Aug 1 '12 at 4:01
@treehau5: $n$ is both the number of people and the number of possibilities. Initially you could have one person knowing each of $0,1,2,\ldots, n-1$ others, which is $n$ possibilities and there would be no ties. Then we argue that you can't have one, decreasing the number of possibilities to $n-1$ –  Ross Millikan Aug 1 '12 at 4:07

First of all, you left out saying that you were proceding by induction, and you left out establishing a base case for the induction.

But let's look at case 2. Suppose that of the original two people, one knows 17 of the 43 people at the party, the other knows 29. Now the newcomer knows the first of the two, but not the other. What makes you think the newcomer knows exactly 29 of the people at the party?

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You are right, I am proceeding by induction, I don't quite understand what you are saying. So I am assuming my n=2, then n+1 comes (the newcommer) in case two n either know each other or don't, but regardless, there will be two people who know the same amount of people at the party. Here n = number of people at the party currently, not some arbitrary number of people they know –  treehau5 Aug 1 '12 at 3:59
If you're assuming $n=2$, then you're only proving it for parties of 3. For your proof to work for parties of 43, you have to show that you can go from any $n$ to $n+1$, not just from 2 to 3. –  Gerry Myerson Aug 1 '12 at 4:06