AMC 12 Mock Exam 4 Problem 24

Number the members $1,\dots,6$ and let $f(x)$ be the leader chosen by member $x$. The rule says exactly that $f(f(x))=f(x)$ for every $x$: the leader $f(x)$ of any member must have chosen herself. So we count functions $f$ from $\{1,\dots,6\}$ to itself with $f(f(x))=f(x)$. Let $S$ be the image of $f$. The condition says $f(y)=y$ for every $y\in S$ (put $y=f(x)$). Conversely, if $f$ fixes every point of its image, then $f(f(x))=f(x)$. So such an $f$ is determined by choosing a nonempty set $S$ of size $k$, letting $f$ be the identity on $S$, and sending each of the other $6-k$ elements anywhere in $S$ (the image is then exactly $S$). Hence the count is $$\sum_{k=1}^{6}\binom6k k^{6-k}=6\cdot1+15\cdot16+20\cdot27+15\cdot16+6\cdot5+1=6+240+540+240+30+1=\boxed{1057}.$$