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

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==Solution==
 
==Solution==
  
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See: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218
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First choose the two letters to be repeated in each set.  <math>5C2=10</math>.  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 <math>10</math> ways to select the first <math>2</math> members and <math>4</math> ways to select the last three, the answer is <math>4*10=40 \rightarrow B</math>.
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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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<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{D}} </math>
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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==
 
==See also==
 
{{AMC10 box|year=2008|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2008|ab=A|num-b=22|num-a=24}}

Revision as of 14:06, 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 $S_{1} = \{a,b,e \}$). 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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