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

Revision as of 18:08, 4 June 2020

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 $1000$ .

(Removed, not a good solution)

Solution 2

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 $\boxed{399}$ .

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 Let <math>A = \{1, 2, 3, 4, 5, 6, 7\}</math>, and let <math>N</math> be the number of functions <math>f</math> from set <math>A</math> to set <math>A</math> such that <math>f(f(x))</math> is a constant function. Find the remainder when <math>N</math> is divided by <math>1000</math>.
-==Solution 1==
+(Removed, not a good solution)
-Let the range be <math>R=\{f(x)|x \in A\}</math>
-Let the constant function be <math>f(f(x))=c</math>, clearly, <math>c \in R</math>
-For every <math>a \in R</math>, since <math>f(f(x))=c</math>, we must have <math>f(a)=c</math>.
-Now we enumerate through possible sizes of <math>R</math>. <math>|R|</math> cannot be more than 4 since all the numbers in <math>R</math> must map to <math>c</math> and any number other than <math>c</math> in <math>R</math> must have at least one number mapped to it.
-. <math>|R|=1</math>
-We have <math>\binom{7}{1}=7</math> choices for <math>R</math>, <math>\binom{1}{1}=1</math> choices for <math>c</math>, and only 1 way to map the 6 numbers that are not in <math>R</math> (remaining numbers) since the range <math>R</math> is size 1. In total we have <math>7\cdot 1\cdot 1=7</math> choices.
-. <math>|R|=2</math>
-We have <math>\binom{7}{2}=21</math> choices for <math>R</math>, <math>\binom{2}{1}=2</math> choices for <math>c</math>.
-Assigning the 5 numbers to the two elements in <math>R</math> yields <math>2^5=32</math> choices. However, one of these choices is to assign all 5 remaining numbers to <math>c</math>, resulting in <math>R</math> being only of size 1. Exclude that and we have <math>32-1=31</math> choices.
-In total we have <math>21*2*31=1302</math> choices.
-. <math>|R|=3</math>
-We have <math>\binom{7}{3}=35</math> choices for <math>R</math>, <math>\binom{3}{1}=3</math> choices for <math>c</math>.
-Notice that assigning the remaining numbers is equivalent to distributing 4 numbers into three groups where two of the three groups must receive at least 1 number (since we already have all numbers in <math>R</math> mapped to <math>c</math>, the restriction is not necessary for the <math>c</math> group). Ignoring the restrictions we have <math>3^4</math> ways. Now minus the two cases where one of the two restricted group is left empty. There are 2 ways to choose the left-out group, and <math>2^4</math> ways to distribute numbers for each choice. Finally, by inclusion-exclusion principle, we need to add back the case in which both restricted groups are left empty, which has only <math>1^4=1</math> occurrence. In all, we have <math>3^4-2*2^4+1^4=50</math> ways to assign the remaining numbers.
-In total, we have <math>35*3*50=5250</math> choices for this case.
-. <math>|R|=4</math>
-We have <math>\binom{7}{4}=35</math> choices for <math>R</math>, <math>\binom{4}{1}=4</math> choices for <math>c</math>.
-Now we have 3 "groups" to fill and only 3 remaining numbers. Thus each group must receive exactly 1 number. However, we still have <math>3!=6</math> ways to permute the mappings.
-Thus, we have <math>35*4*6=840</math> choices in total.
-To summarize, we have <math>7+1302+5250+840=7399</math> different mappings possible. So the answer is <math>\boxed{399}</math>.
 ==Solution 2==

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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Difference between revisions of "2013 AIME II Problems/Problem 11"

Revision as of 18:08, 4 June 2020

Problem 11

Solution 2

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