Difference between revisions of "2008 AMC 10A Problems/Problem 23"
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==Solution=== | ==Solution=== | ||
| − | + | 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>. | |
==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 19:05, 15 June 2008
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. 
. 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 
ways to select the first 
members and 
ways to select the last three, the answer is 
.
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 | ||
