# Boys dancing with girls

In a party there are $r$ boys and $m$ girls. The first boy dances with $5$ girls, the second boy dances with $6$ girls, and the last boy dances with all girls. What is the relation between $r$ and $m$? I managed to find that $m=r+4,$ but it was not using combinatorics.

Let me explain how I found that answer: first boy dances with $a_{1}$ girls, second boy dances with $a_{2}$girls, last boy dances with $a_{r}$ girls. Then $5=a_{1}=1+4, 6=a_{2}=2+4,\cdots, m=a_{r}=4+r.$

My question: How does one solve this question by using combinatorics?

-
The question is not well formulated, since nothing is said about the boys after the second but before the last. You probably mean that every next boy dances with one more girl than the previous boy, up to and including the last boy, because without stating this trend there is really not much you can say about $r$ and $m$. – Marc van Leeuwen Aug 9 at 14:27

What you've done is fine, if a bit informal. You can make it a bit nicer as follows. Clearly the intended pattern is that each succeeding boy dances with one more girl. If $a_k$ is the number of girls whom the $k$-th boy dances with, then you have $a_1=5,a_2=6,\dots,a_r=m$. In other words, $\langle 5,6,\dots m\rangle$ is supposed to be a sequence of $r$ consecutive integers. This sequence has $m-4$ terms, so we must have $r=m-4$.

-

Your argument is combinatorial. Here is a variant.

Bring in $4$ more boys from the other class, the first to dance with $1$ girl, the second with $2$ girls, the third with $3$, the fourth with $4$, and let the rest of the action be as described. Then the number of boys would be the same as the number of girls, namely $m$. Now tell the $4$ new boys to go away. So we are left with $m-4$, which is $r$.

-