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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?

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2 Answers 2

up vote 3 down vote accepted

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 $r=m-4$.

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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 $r$.

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