2008 AMC 10A Problems/Problem 23
Problem
Two subsets of the set 
are to be chosen so that their union is 
and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
Solution
First choose the two letters to be repeated in each set. 
. Now we have three remaining elements that we wish to place into two separate subsets. There are 
ways to do so (Do you see why? It's because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted (Take for example 
and 
). Notice how 
and 
are interchangeable. A simple division by two will fix this problem. Thus we have:
Alternatively, after picking the two elements in both sets in ways, we can use stars and bars to assign the remaining 3 elements to the sets. There are 3 stars, and 1 bar, so there are 4 total ways of assigning the elements. Then there are 
ways to create the sets.
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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