1993 AIME Problems/Problem 8
Contents
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 (Overcounting)
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 (Overcounting)
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 the 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: Note that for all each unordered pair of the -element set contributes three unordered pairs to the -element set, as the element can be added to either or both of the subsets. The only exception is that the unordered pair of identical subsets of the -element set, namely only contributes two unordered pairs to the -element set. The table above shows this relationship between the -element set and the -element set.
Together, the recursive formula for is Two solutions follow from here:
Solution 3.1 (Recursive Formula)
We evaluate recursively: ~MRENTHUSIASM
Solution 3.2 (Explicit Formula)
For all we have which resembles the result in Solutions 1 and 2. The answer is
As note that holds for too. Therefore, this formula is true for all nonnegative integers
~MRENTHUSIASM
Solution 4 (Casework)
We can perform casework based on the number of overlapping elements. If no elements overlap, there is way to choose the overlapping elements, and ways to distribute the remaining elements--each element can go in one subset or the other. We must also divide by because the order of the subsets does not matter. Proceeding similarly for the other cases, our sum is (Note that we have to be especially careful with the last case, as it does not follow the pattern of the other cases.) Adding these up gives a total of
~vaporwave
~MrThinker (LaTeX improvements)
Solution 5 (Bash)
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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.