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2008 AMC 10A Problems/Problem 23

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Problem

Two subsets of the set S=\lbrace a,b,c,d,e\rbrace are to be chosen so that their union is S 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?

\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320

Solution

First choose the two letters to be repeated in each set. \dbinom{5}{2}=10. Now we have three remaining elements that we wish to place into two separate subsets. There are 2^3 = 8 ways to do so (Do you see why?). Unfortunately, we have over-counted (Take for example S_{1} = \{a,b,c,d \} and S_{2} = \{a,b,e \}). Notice how S_{1} and S_{2} are interchangeable. A simple division by two will fix this problem. Thus we have:

\dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{D}}

This problem was discussed in an AoPS Math Jam a while back. The transcript should be located here: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218

See also

2008 AMC 10A (ProblemsResources)
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
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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