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

See: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218

First choose the two letters to be repeated in each set. $5C2=10$. Then we have three distinguishable elements to put into two indistinguishable groups. We can distribute these elements as all in one group, or as one in the first group and two in the second group. There is one way to put all the remaining elements into one of the indistinguishable groups. There are three ways to separate one member from the other two. Since there are $10$ ways to select the first $2$ members and $4$ ways to select the last three, the answer is $4*10=40 \rightarrow B$.

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions