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Difference between revisions of "2013 AIME II Problems/Problem 11"

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Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>.
 
Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>.
  
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==Solution 2==
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==Solution 1 Clarified==
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Define the three layers as [domain] <math>x</math>, [codomain] <math>f(x)</math>, and codomain <math>f(f(x))</math>. Each one of them is contained in the set <math>A</math>. We know that <math>f(f(x))</math> is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value <math>c</math> in <math>7</math> ways. So now, we choose the values that can be <math>f(x)</math> for all those values should satisfy <math>f(f(x))=c</math>. Let's <math>S</math> be that set of values. First things first, we must have <math>5</math> to be part of <math>S</math>, for the <math>S</math> is part of the domain of <math>x</math>. Since the values in <math>i\in S</math> all satisfy <math>f(i) = 5</math>, we have <math>5</math> to be a value that <math>f(x)</math> can be. Now, for the elements other than <math>5</math>:
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If we have <math>k</math> elements other than <math>5</math> than can be part of <math>S</math>, we will have <math>\binom{6}{k}</math> ways to choose those values. There will also be <math>k</math> ways for each of the elements in <math>A</math> other than <math>5</math> and those in set <math>S</math> (for when function <math>f</math> is applied on those values, we already know it would be <math>5</math>). There are <math>6-k</math> elements in <math>A</math> other than <math>5</math> and those in set <math>S</math>. So, there should be <math>6^{6-k}</math> ways to match the domain <math>x</math> to the values of <math>f(x)</math>. So, summing up all possible values of <math>k</math> (<math>[1,6]</math>), we have
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<cmath>\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057</cmath>
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Multiplying that by the original <math>7</math> for the choice of <math>c</math>, we have <math>7 \cdot 1057 = 7\boxed{399}.</math>
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~sml1809
  
 
==See Also==
 
==See Also==

Revision as of 14:53, 8 January 2023

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 $\boxed{399}$.

Solution 1 Clarified

Define the three layers as [domain] $x$, [codomain] $f(x)$, and codomain $f(f(x))$. Each one of them is contained in the set $A$. We know that $f(f(x))$ is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value $c$ in $7$ ways. So now, we choose the values that can be $f(x)$ for all those values should satisfy $f(f(x))=c$. Let's $S$ be that set of values. First things first, we must have $5$ to be part of $S$, for the $S$ is part of the domain of $x$. Since the values in $i\in S$ all satisfy $f(i) = 5$, we have $5$ to be a value that $f(x)$ can be. Now, for the elements other than $5$:

If we have $k$ elements other than $5$ than can be part of $S$, we will have $\binom{6}{k}$ ways to choose those values. There will also be $k$ ways for each of the elements in $A$ other than $5$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $5$). There are $6-k$ elements in $A$ other than $5$ and those in set $S$. So, there should be $6^{6-k}$ ways to match the domain $x$ to the values of $f(x)$. So, summing up all possible values of $k$ ($[1,6]$), we have

\[\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057\]

Multiplying that by the original $7$ for the choice of $c$, we have $7 \cdot 1057 = 7\boxed{399}.$

~sml1809

See Also

2013 AIME II (Problems • Answer Key • Resources)
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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