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 "2010 AIME II Problems/Problem 8"

(→‎Solution 3 (PIE and Complementary Counting))
(→‎Solution 3 (PIE and Complementary Counting))
 
(10 intermediate revisions by the same user not shown)
Line 34: Line 34:
 
== Solution 3 (PIE and Complementary Counting) ==
 
== Solution 3 (PIE and Complementary Counting) ==
  
−
The total number of possible subsets is <math>\sum_{i=1}^{11}\dbinom{12}{i}</math>, which is <math>2^{12}-2</math>. Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude <math>0</math> and <math>12</math> since they leave a set empty. We proceed with complementary counting and casework:
+
The total number of possible subsets is <math>\sum_{i=1}^{11}\dbinom{12}{i}</math>, which is <math>2^{12}-2</math>. Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude <math>i=0</math> and <math>i=12</math> since they leave a set empty. We proceed with complementary counting and casework:
  
−
We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where the element #elements in <math>A</math> is in <math>A</math>, which violates the first condition. Then we find the ways where the elements #elements in <math>B</math> and <math>12</math>-#elements in <math>B</math> are in set <math>B</math>, which violates only the second condition, and not the first (If we did, we would overcount the union of both cases). This ensures all possible cases are covered.  
+
We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where <math>|A|</math> is in <math>A</math>, which violates the first condition. Then we find the ways where the elements <math>|B|</math> and <math>12-|B|</math> are in set <math>B</math>, which violates only the second condition, and not the first. This ensures we do not overcount.
  
−
Case 1: #elements in <math>A</math> is an element in <math>A</math>
+
Case 1: <math>|A|</math> is an element in <math>A</math>
  
 
There are <math>\sum_{i=1}^{11}\dbinom{11}{i-1}</math> = <math>2^{11}-1</math> ways in this case.
 
There are <math>\sum_{i=1}^{11}\dbinom{11}{i-1}</math> = <math>2^{11}-1</math> ways in this case.
  
−
Case 2: #elements in <math>B</math> and 12-#elements are in <math>B</math>
+
Case 2: <math>|B|</math> and <math>12-|B|</math> are in <math>B</math>
  
−
We introduce a subcase where #elements in <math>B</math> is not <math>6</math>, since it would also result in <math>12-6=6</math> for the other element.
+
We introduce a subcase where <math>|B|</math> is not 6, since the other element would also be 6.
−
There are <math>B-2</math> items to choose from <math>12-2=10</math>, the <math>2</math> being the <math>2</math> mandatory elements in <math>B</math>. Therefore, <math>B</math> can be from <math>2</math> to <math>11</math>.
+
There are <math>B-2</math> elements to choose from 10 elements. Therefore, <math>|B|</math> can be from 2 to 11.
 
There are <math>\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211</math> ways.
 
There are <math>\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211</math> ways.
−
The other subcase is when #elements in <math>B</math> are equal to <math>6</math>. There are <math>\dbinom{11}{5}</math> ways for this case, which equals <math>462</math>.
+
The other subcase is when <math>|B|</math> is equal to 6. There are <math>\dbinom{11}{5}=462</math> ways.
 
Adding the subcases gives us <math>2^{10}+251</math>.  
 
Adding the subcases gives us <math>2^{10}+251</math>.  
  

Latest revision as of 19:57, 20 July 2023

Problem

Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:

  • $\mathcal{A} \cup \mathcal{B} = \{1,2,3,4,5,6,7,8,9,10,11,12\}$,
  • $\mathcal{A} \cap \mathcal{B} = \emptyset$,
  • The number of elements of $\mathcal{A}$ is not an element of $\mathcal{A}$,
  • The number of elements of $\mathcal{B}$ is not an element of $\mathcal{B}$.

Find $N$.

Solution

Let us partition the set $\{1,2,\cdots,12\}$ into $n$ numbers in $A$ and $12-n$ numbers in $B$,

Since $n$ must be in $B$ and $12-n$ must be in $A$ ($n\ne6$, we cannot partition into two sets of 6 because $6$ needs to end up somewhere, $n\ne 0$ or $12$ either).

We have $\dbinom{10}{n-1}$ ways of picking the numbers to be in $A$.

So the answer is $\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}$.

Note: We have $\dbinom{10}{n-1}$ ways of picking the numbers to be in $A$ because there are $n$ numbers in $A$ and since $12-n$ is already a term in the set we simply have to choose another $n-1$ numbers from the $10$ numbers that are available.

Solution 2

Regardless of the size $n$ of $A$ (ignoring the case when $n = 6$), $n$ must not be in $A$ and $12 - n$ must be in $A$.

There are $10$ remaining elements whose placements have yet to be determined. Note that the actual value of $n$ does not matter; there is always $1$ necessary element, $1$ forbidden element, and $10$ other elements that need to be distributed. There are $2$ places to put each of these elements, for $2^{10}$ possibilities.

However, there is the edge case of $n = 6; 6$ is forced not the be in either set, so we must subtract the $\dbinom{10}{5}$ cases where $A$ and $B$ have size $6$.

Thus, our answer is $2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}$


Solution 3 (PIE and Complementary Counting)

The total number of possible subsets is $\sum_{i=1}^{11}\dbinom{12}{i}$, which is $2^{12}-2$. Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude $i=0$ and $i=12$ since they leave a set empty. We proceed with complementary counting and casework:

We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where $|A|$ is in $A$, which violates the first condition. Then we find the ways where the elements $|B|$ and $12-|B|$ are in set $B$, which violates only the second condition, and not the first. This ensures we do not overcount.

Case 1: $|A|$ is an element in $A$

There are $\sum_{i=1}^{11}\dbinom{11}{i-1}$ = $2^{11}-1$ ways in this case.

Case 2: $|B|$ and $12-|B|$ are in $B$

We introduce a subcase where $|B|$ is not 6, since the other element would also be 6. There are $B-2$ elements to choose from 10 elements. Therefore, $|B|$ can be from 2 to 11. There are $\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211$ ways. The other subcase is when $|B|$ is equal to 6. There are $\dbinom{11}{5}=462$ ways. Adding the subcases gives us $2^{10}+251$.

Adding this with case one gives us $2^{11}+2^{10}+250$. Subtracting this from $2^{12}-2$ gives $1024-252=\boxed{772}$.

~Magnetoninja

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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=2010_AIME_II_Problems/Problem_8&oldid=195931"