Difference between revisions of "2002 AIME II Problems/Problem 9"
I like pie (talk | contribs) (Added problem, solution still needed) |
(I found something to compute, I just don't know how to compute it.) |
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== Solution == | == Solution == | ||
| + | Let's say that the first set has <math>x</math> elements and the second set has <math>y</math> elements, where <math>x+y\leq 10</math>. The number of <math>\mathcal{S}</math> is <math>\binom{10}{x}\binom{10-x}{y}=\dfrac{10!(10-x)!}{x!(10-x)!y!(10-x-y)!}=\dfrac{10!}{x!y!(10-x-y)!}</math> | ||
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| + | We now divide by 2, since order doesn't matter in <math>\mathcal{S}</math>: <math>\dfrac{5\cdot 9!}{x!y!(10-x-y)!}</math>. Thus | ||
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| + | <cmath>n=\sum_{x+y\leq 10} \dfrac{5\cdot 9!}{x!y!(10-x-y)!}.</cmath> | ||
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{{solution}} | {{solution}} | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=8|num-a=10}} | {{AIME box|year=2002|n=II|num-b=8|num-a=10}} | ||
Revision as of 13:52, 19 April 2008
Problem
Let 
be the set 
Let 
be the number of sets of two non-empty disjoint subsets of . (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when is divided by 
.
Solution
Let's say that the first set has 
elements and the second set has 
elements, where 
. The number of is 
We now divide by 2, since order doesn't matter in : 
. Thus
![\[n=\sum_{x+y\leq 10} \dfrac{5\cdot 9!}{x!y!(10-x-y)!}.\]](http://latex.artofproblemsolving.com/6/b/c/6bc485dc805d993fa2245324b1dd87eba0fdb660.png)
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 2002 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
