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Difference between revisions of "2008 AMC 10A Problems/Problem 23"

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First choose the two letters to be repeated in each set.  <math>\dbinom{5}{2}=10</math>.  Now we have three remaining elements that we wish to place into two separate subsets. There are <math>2^3 = 8 </math> ways to do so (Do you see why?). Unfortunately, we have over-counted  (Take for example <math>S_{1} = \{a,b,c,d \}</math> and <math>S_{1} = \{a,b,e \}</math>). Notice how <math>S_{1}</math> and <math>S_{2}</math> are interchangeable. A simple division by two will fix this problem. Thus we have:
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First choose the two letters to be repeated in each set.  <math>\dbinom{5}{2}=10</math>.  Now we have three remaining elements that we wish to place into two separate subsets. There are <math>2^3 = 8 </math> ways to do so (Do you see why?). Unfortunately, we have over-counted  (Take for example <math>S_{1} = \{a,b,c,d \}</math> and <math>S_{2} = \{a,b,e \}</math>). Notice how <math>S_{1}</math> and <math>S_{2}</math> are interchangeable. A simple division by two will fix this problem. Thus we have:
  
 
<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{D}} </math>
 
<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{D}} </math>

Revision as of 13:07, 11 August 2009

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 (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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