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

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 who’s 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, this ignores the case when $n = 6; 6$ 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, out answer is $2^10 - \dbinom{10}{5} = 1024 - 252 = \boxed{772}$

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

See also