Tuesday, February 9, 2016

Dancing partners

In a party with $n$ boys and $n$ girls, each boy dances with some girls and vice versa. (No pair of boy and girl dance more than once). There are $n^2-n+1$ dances that occurred throughout the night. Show that we can pair each boy with each girl so that everyone is paired with someone he/she danced with.

Solution

The key fact is because the number of dances is more than $n^2-n$. If it were exactly $n^2-n$, we could have a situation where $n-1$ boys danced with all the girls, and one boy never danced, and this effectively blocks that boy from being paired correctly.

Suppose that there is no way to pair them correctly. That is, for every given pairing, there's always a pair that did not dance. For each pairing, we define the score of said pairing to be the number of pairs that actually danced. This means a legal pairing is the one with score $n$.

Now let $S$ be the sum of all the scores of all possible pairings. Our contrapositive hypothesis is that all pairings have a score of at most $n-1$, so $S \leq (n-1)n!$ (because there are $n!$ possible pairings).

On the other hand, for each dance that happened throughout the night, we want to count how many points it contributes to $S$. Suppose the dance happens between boy $B$ and girl $G$. Then it contributes one point of score whenever the pairing pairs $B$ and $G$. In other words, there are $(n-1)!$ such pairings (because one pair is fixed and we have complete freedom to pair the other $n-1$ boys and $n-1$ girls. So the total number of score is: $$S = (n^2-n+1)(n-1)! > (n^2-n)(n-1)! = (n-1)n!$$ A contradiction.