2010 AIME II Problems/Problem 8: Difference between revisions

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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 it would also result in $12-6=6$ for the other element. 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

@@ Line 34: / Line 34: @@
 == 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 <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 (If we did, we would overcount the intersection 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: <math>|A|</math> is an element in <math>A</math>
+Case <math>1</math>: <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.
-Case 2: <math>|B|</math> and <math>12-|B|</math> are in <math>B</math>
+Case <math>2</math>: <math>|B|</math> and <math>12-|B|</math> are in <math>B</math>
 We introduce a subcase where <math>|B|</math> is not <math>6</math>, since it would also result in <math>12-6=6</math> for the other element.
-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 <math>10</math> elements. Therefore, <math>|B|</math> can be from <math>2</math> to <math>11</math>.
 There are <math>\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211</math> ways.
 The other subcase is when <math>|B|</math> is equal to <math>6</math>. There are <math>\dbinom{11}{5}=462</math> ways.

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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