2013 AIME II Problems/Problem 11

Problem 11

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.

Solution

Any such function can be constructed by distributing the elements of $A$ on three tiers.

The bottom tier contains the constant value, $c=f(f(x))$ for any $x$. (Obviously $f(c)=c$.)

The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$, where $1\le k\le 6$.

The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier.

There are $7$ choices for $c$. Then for a given $k$, there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier.

Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$, giving the answer $\boxed{399}$.

See Also

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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