2010 AIME II Problems/Problem 8

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 $0$ and $12$ 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 the element #elements in $A$ is in $A$ , which violates the first condition. Then we find the ways where the elements #elements in $B$ and 12-#elements in $B$ are in set $B$ , 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.

Case 1: #elements in $A$ is an element in $A$

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

Case 2: #elements in $B$ and 12-#elements are in $B$

We introduce a subcase where #elements in $B$ is not $6$ , since it would also result in $12-6=6$ for the other element. There are $B-2$ items to choose from $12-2=10$ , the $2$ being the $2$ mandatory elements in $B$ . 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 #elements in $B$ are equal to $6$ . There are $\dbinom{11}{5}$ ways for this case, which equals $462$ . 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 $1024-252=772$ .

-Magnetoninja

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2010 AIME II Problems/Problem 8

Contents

Problem

Solution

Solution 2

Solution 3 (PIE and Complementary Counting)

See also