Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AIME II Problems/Problem 11"
m (→‎Solution)
Line 13: Line 13:
 
1. <math>|R|=1</math>
 
1. <math>|R|=1</math>
  
−
We have <math>\binom{1}{7}=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.  
+
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.  
  
 
2. <math>|R|=2</math>
 
2. <math>|R|=2</math>
  
−
We have <math>\binom{2}{7}=21</math> choices for <math>R</math>, <math>\binom{1}{2}=2</math> choices for <math>c</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.  
 
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.  
Line 25: Line 25:
 
3. <math>|R|=3</math>
 
3. <math>|R|=3</math>
  
−
We have <math>\binom{3}{7}=35</math> choices for <math>R</math>, <math>\binom{1}{3}=3</math> choices for <math>c</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.  
 
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.  
Line 33: Line 33:
 
4. <math>|R|=4</math>
 
4. <math>|R|=4</math>
  
−
We have <math>\binom{4}{7}=35</math> choices for <math>R</math>, <math>\binom{1}{4}=4</math> choices for <math>c</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.  
 
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.  
Line 39: Line 39:
 
Thus, we have <math>35*4*6=840</math> choices in total.  
 
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>.  
+
To summarize, we have <math>7+1302+5250+840=7399</math> different mappings possible. So the answer is <math>\boxed{399}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:05, 31 December 2013

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

Let the range be $R=\{f(x)|x \in A\}$

Let the constant function be $f(f(x))=c$, clearly, $c \in R$

For every $a \in R$, since $f(f(x))=c$, we must have $f(a)=c$.

Now we enumerate through possible sizes of $R$. $|R|$ cannot be more than 4 since all the numbers in $R$ must map to $c$ and any number other than $c$ in $R$ must have at least one number mapped to it.

1. $|R|=1$

We have $\binom{7}{1}=7$ choices for $R$, $\binom{1}{1}=1$ choices for $c$, and only 1 way to map the 6 numbers that are not in $R$ (remaining numbers) since the range $R$ is size 1. In total we have $7\cdot 1\cdot 1=7$ choices.

2. $|R|=2$

We have $\binom{7}{2}=21$ choices for $R$, $\binom{2}{1}=2$ choices for $c$.

Assigning the 5 numbers to the two elements in $R$ yields $2^5=32$ choices. However, one of these choices is to assign all 5 remaining numbers to $c$, resulting in $R$ being only of size 1. Exclude that and we have $32-1=31$ choices.

In total we have $21*2*31=1302$ choices.

3. $|R|=3$

We have $\binom{7}{3}=35$ choices for $R$, $\binom{3}{1}=3$ choices for $c$.

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 $R$ mapped to $c$, the restriction is not necessary for the $c$ group). Ignoring the restrictions we have $3^4$ 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 $2^4$ 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 $1^4=1$ occurrence. In all, we have $3^4-2*2^4+1^4=50$ ways to assign the remaining numbers.

In total, we have $35*3*50=5250$ choices for this case.

4. $|R|=4$

We have $\binom{7}{4}=35$ choices for $R$, $\binom{4}{1}=4$ choices for $c$.

Now we have 3 "groups" to fill and only 3 remaining numbers. Thus each group must receive exactly 1 number. However, we still have $3!=6$ ways to permute the mappings.

Thus, we have $35*4*6=840$ choices in total.

To summarize, we have $7+1302+5250+840=7399$ different mappings possible. So the answer is $\boxed{399}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_11&oldid=58456"