More Is Less

So, I find it to be rather unfortunate that the night after posting my puzzle about shaking hands from last time, I managed to come up with a similar problem that I consider to be more interesting. This leaves me with the issue of either solving that puzzle from last time and then posting the new puzzle as a follow-up, or just posting the new puzzle and solving them both at once next time. The problem with the former is that after the first puzzle is solved, the second puzzle is very easy to do (its the exact same logic), but it does have one extra thing one needs to notice in order to solve it, so it still is not trivial. The problem with the latter is that I find it unappealing for aesthetic reasons that don't really make sense.

Going with aesthetics, I guess I might as well solve the puzzle from last time now. The logic is very simple, first of all notice that each person has 2N people available to shake hands with (of the 2N+2 people, they cannot shake their own or their partners), and so the number of hands each person shakes is between 0 and 2N. You asked 2N+1 people how many hands they shook, and you received 2N+1 unique answers, thus you must have received every answer between 0 and 2N, pigenhole principle and all that. We will identify each person by the number of hands they shook, that is, person k shook k hands, and this assignment gives each person a unique number.

Person 2N must have shaken hands with every person they could, and person 0 must have done the opposite. These two people must be partners, or those two people did not go to the same party. Thus, (0,2N) is a pair. Now, each person at the party outside of that pair must have shaken hands with exactly one person in that pair (2N, to be specific), so lets just remove pair (0,2N) and reduce everybodys number by 1. This has now set up the exact same situation as the original problem, but we have reduced N to N-1. Everybody's number is 1 lower, but they are all still unique. This means that whatever the solution is to the problem with N (call that F(N)) it will be 1 lower when N is instead N-1. That is to say
F(N) = F(N-1)+1

Which means
F(N)=F(1)+N-1

so we just have to solve the N=1 case.

When N=1, the answers you were told were 0,1, and 2. 2 and 0 must be partners by the same logic as before, and 1 must be your own partner. Since 2 shook hands with 2 people, the other hand must have been yours, and 0 did not shake your hand. This means that F(1) is 1 and therefore F(N) is N. Proving that you shook hands with N people, exactly 1 person from each couple.

Now for the follow-up puzzle:
You go to a party with 2N other people (so there are 2N+1 people in total at this party). At the party people shake hands with other people. No person shakes hands with themselves. After the party, you ask each person that was at the party how many different people they shook hands with, and each person gives you a unique answer. How many different people did you shake hands with?

Very much the same setup, but the "couples" concept has been stripped away.

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