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 $\boxed{399}$ .

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2013 AIME II Problems/Problem 11

Contents

Problem 11

Solution

Solution 2

See Also