math-combo-01

# Counting Idempotent Self-Maps

Let $A = \{1, 2, 3, 4, 5\}$. Count the number of functions $f : A \to A$ that satisfy
$$f(f(x)) = f(x) \qquad \text{for every } x \in A.$$

(Such a function acts as the identity on its own range.)

Give your answer as a single integer.

I need to count the number of functions $f : A \to A$ where $A = \{1, 2, 3, 4, 5\}$ such that $f(f(x)) = f(x)$ for all $x \in A$.

Let me think about what the condition $f(f(x)) = f(x)$ means.

For any element $y$ in the range of $f$, we have $f(y) = y$. This is because if $y = f(x)$ for some $x$, then $f(f(x)) = f(x)) = y$, and since $f(f(x)) = f(x)$, we get $f(y) = y$.

So any function satisfying $f(f(x)) = f(x)$ must be the identity on its range.

Let me approach this systematically by considering what happens to elements under $f$.

For each element $x \in A$, either:
1. $f(x) = x$ (fixed point)
2. $f(x) = y$ where $y \neq x$ and $f(y) = y$ (so $y$ is a fixed point)

But wait, let me reconsider this more carefully.

If $f(f(x)) = f(x)$, then for any element $y$ in the range of $f$, we have $f(y) = y$. This means that every element in the image of $f$ must be a fixed point of $f$.

So I can think of this as:
- Choose a subset $S \subseteq A$ to be the set of fixed points
- For each element not in $S$, map it to some element in $S$
- The elements in $S$ must map to themselves

Let me count by considering all possible subsets $S$ of $A$ that could serve as the set of fixed points.

For any subset $S \subseteq A$, I can define a function $f$ such that:
- For all $x \in S$: $f(x) = x$
- For all $x \notin S$: $f(x) \in S$

The number of such functions is the number of ways to choose a subset $S$ and then assign each element not in $S$ to an element in $S$.

If $|S| = k$, then there are $\binom{5}{k}$ ways to choose $S$, and for the remaining $(5-k)$ elements, there are $k^{5-k}$ ways to map them to elements of $S$.

So the total number of functions is:
$$\sum_{k=0}^{5} \binom{5}{k} k^{5-k}$$

Let me compute each term:

For $k = 0$: $\binom{5}{0} 0^5 = 1 \cdot 0 = 0$

For $k = 1$: $\binom{5}{1} 1^4 = 5 \cdot 1 = 5$

For $k = 2$: $\binom{5}{2} 2^3 = 10 \cdot 8 = 80$

For $k = 3$: $\binom{5}{3} 3^2 = 10 \cdot 9 = 90$

For $k = 4$: $\binom{5}{4} 4^1 = 5 \cdot 4 = 20$

For $k = 5$: $\binom{5}{5} 5^0 = 1 \cdot 1 = 1$

Total: $0 + 5 + 80 + 90 + 20 + 1 = 196$

Wait, let me double-check this interpretation. 

Actually, there's a more direct way to think about it.

If $f(f(x)) = f(x)$ for all $x$, then:
- The function $f$ is idempotent
- The range of $f$ consists entirely of fixed points
- For any element in the range, applying $f$ gives itself

This is equivalent to choosing a subset $S \subseteq A$ and defining $f$ such that:
- $f(x) = x$ for all $x \in S$
- $f(x) \in S$ for all $x \notin S$

The number of such functions equals the number of ways to choose a subset $S$ (the fixed points) and then map the remaining elements into $S$.

This is exactly what I computed above.

$\boxed{196}$