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_{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:
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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? 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 <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{B}} </math>
 
<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{B}} </math>

Revision as of 00:19, 23 November 2011

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:

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

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
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