Difference between revisions of "2010 AIME II Problems/Problem 8"
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== 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 | + | 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: | + | 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: | + | Case 2: <math>|B|</math> and <math>12-|B|</math> are in <math>B</math> |
− | We introduce a subcase where | + | 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> | + | 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 | + | 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
Contents
[hide]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 .
Note: We have ways of picking the numbers to be in because there are numbers in and since is already a term in the set we simply have to choose another numbers from the numbers that are available.
Solution 2
Regardless of the size of (ignoring the case when ), must not be in and must be in .
There are remaining elements whose 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
Solution 3 (PIE and Complementary Counting)
The total number of possible subsets is , which is . Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude and 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 is in , which violates the first condition. Then we find the ways where the elements and are in set , which violates only the second condition, and not the first. This ensures we do not overcount.
Case 1: is an element in
There are = ways in this case.
Case 2: and are in
We introduce a subcase where is not 6, since the other element would also be 6. There are elements to choose from 10 elements. Therefore, can be from 2 to 11. There are ways. The other subcase is when is equal to 6. There are ways. Adding the subcases gives us .
Adding this with case one gives us . Subtracting this from gives .
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 |
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