Difference between revisions of "1993 AIME Problems/Problem 8"
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== Solution 3 (Recursion) == | == Solution 3 (Recursion) == | ||
− | For all nonnegative integers <math>n,</math> let <math>n</math> be the number of elements in <math>S,</math> and <math>f(n)</math> be the number of unordered pairs | + | For all nonnegative integers <math>n,</math> let <math>n</math> be the number of elements in <math>S,</math> and <math>f(n)</math> be the number of unordered pairs <math>\{A,B\}</math> of subsets of <math>S</math> for which <math>A\cup B=S.</math> We wish to find <math>f(6).</math> |
Without the loss of generality, let the elements of <math>S</math> be <math>1,2,3,\cdots,n</math> Based on the value of <math>n,</math> we construct the following table: | Without the loss of generality, let the elements of <math>S</math> be <math>1,2,3,\cdots,n</math> Based on the value of <math>n,</math> we construct the following table: |
Revision as of 02:21, 2 September 2021
Contents
[hide]Problem
Let be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of so that the union of the two subsets is ? The order of selection does not matter; for example, the pair of subsets represents the same selection as the pair
Solution 1
Call the two subsets and For each of the elements in we can assign it to either or both. This gives us possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both and contain all elements of So our final answer is then
Solution 2
Given one of subsets with elements, the other also has possibilities; this is because it must contain all of the "missing" elements and thus has a choice over the remaining We want by Binomial Theorem. But the order of the sets doesn't matter, so we get
Solution 3 (Recursion)
For all nonnegative integers let be the number of elements in and be the number of unordered pairs of subsets of for which We wish to find
Without the loss of generality, let the elements of be Based on the value of we construct the following table:
Solution 4 (Extremely Bashy)
If we wanted to, we could use casework, being very careful not to double count cases. Let's first figure out what cases we need to look at. The notation I will be using is which implies that we pick numbers for the first set which then the second set can have numbers.
Clearly: However notice that many of the cases are double counted as direction does not matter, e.g. is the same as Get rid of those cases so we are just left with: Now we start computing, is just case, is just cases, is just cases, and is just cases (If you have trouble understanding this, write down the six letters and then try to understand what really means.). Now what you can do is continue on this same pattern like Solution 2 does, and then use simple symmetry to figure out the double counted cases. However, the purpose of this solution is to bash out the double counted cases, so we will do exactly that.
One quick thing though. We have a double counted case with the as choosing ABC and DEF is the same thing as choosing DEF and then ABC. There are cases of this.
For computing we use the same process as before. We have (Note, the comes from ), and for we have and then for we just have (there is no double counted case since ABCDEF, ABCDEF is only counted once).
Summing case by case, we have
Disclaimer: This is a terrible solution, it should only be done as practice for casework bashing and never should be done on a real AIME.
~Tost (Solution)
~MRENTHUSIASM (Reformatting)
See also
1993 AIME (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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