Difference between revisions of "2010 AIME II Problems/Problem 8"
m (→Solution 2) |
(→Solution 2) |
||
Line 25: | Line 25: | ||
There are <math>10</math> remaining elements who’s placements have yet to be determined. Note that the actual value of <math>n</math> does not matter; there is always <math>1</math> necessary element, <math>1</math> forbidden element, and <math>10</math> other elements that need to be distributed. There are <math>2</math> places to put each of these elements, for <math>2^{10}</math> possibilities. | There are <math>10</math> remaining elements who’s placements have yet to be determined. Note that the actual value of <math>n</math> does not matter; there is always <math>1</math> necessary element, <math>1</math> forbidden element, and <math>10</math> other elements that need to be distributed. There are <math>2</math> places to put each of these elements, for <math>2^{10}</math> possibilities. | ||
− | However, | + | However, there is the edge case of <math>n = 6; 6</math> is forced not the be in either set, so we must subtract the <math>\dbinom{10}{5}</math> cases where <math>A</math> and <math>B</math> have size <math>6</math>. |
Thus, our answer is <math>2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}</math> | Thus, our answer is <math>2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}</math> |
Revision as of 14:39, 6 January 2020
Contents
Problem
Let be the number of ordered pairs of nonempty sets and that have the following properties:
- ,
- ,
- The number of elements of is not an element of ,
- The number of elements of is not an element of .
Find .
Solution
Let us partition the set into numbers in and numbers in ,
Since must be in and must be in (, we cannot partition into two sets of 6 because needs to end up somewhere, or either).
We have ways of picking the numbers to be in .
So the answer is .
Solution 2
Regardless of the size of (ignoring the case when ), must not be in and must be in .
There are remaining elements who’s placements have yet to be determined. Note that the actual value of does not matter; there is always necessary element, forbidden element, and other elements that need to be distributed. There are places to put each of these elements, for possibilities.
However, there is the edge case of is forced not the be in either set, so we must subtract the cases where and have size .
Thus, our answer is
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.