2008 AMC 10A Problems/Problem 23

Revision as of 19:54, 3 January 2013 by Pl210741 (talk | contribs) (Solution)

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? 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 $S_{1} = \{a,b,c,d \}$ 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:

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


Alternatively, after picking the two elements in both sets in $\dbinom{5}{2}=10$ 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 $10\cdot4=4-$ ways to create the sets.

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