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: | + | 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 23:19, 22 November 2011
Problem
Two subsets of the set 
are to be chosen so that their union is 
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?
Solution
First choose the two letters to be repeated in each set. 
. Now we have three remaining elements that we wish to place into two separate subsets. There are 
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 
and 
). Notice how 
and 
are interchangeable. A simple division by two will fix this problem. Thus we have:
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 | ||
