Difference between revisions of "1989 AIME Problems/Problem 2"

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== Problem ==
 
== Problem ==
Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices?
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Ten [[point]]s are marked on a [[circle]]. How many distinct [[convex polygon]]s of three or more sides can be drawn using some (or all) of the ten points as [[vertex | vertices]]?
  
 
== Solution ==
 
== Solution ==
{{solution}}
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Any [[subset]] of the ten points with three or more members can be made into exactly one such polygon.  Thus, we need to count the number of such subsets.  There are <math>2^{10} = 1024</math> total subsets of a ten-member [[set]], but of these <math>{10 \choose 0} = 1</math> have 0 members, <math>{10 \choose 1} = 10</math> have 1 member and <math>{10 \choose 2} = 45</math> have 2 members.  Thus the answer is <math>1024 - 1 - 10 - 45 = 968</math>.
  
 
== See also ==
 
== See also ==
* [[1989 AIME Problems/Problem 3|Next Problem]]
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{{AIME box|year=1989|num-b=1|num-a=3}}
* [[1989 AIME Problems/Problem 1|Previous Problem]]
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* [[1989 AIME Problems]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 10:08, 25 February 2007

Problem

Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices?

Solution

Any subset of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are $2^{10} = 1024$ total subsets of a ten-member set, but of these ${10 \choose 0} = 1$ have 0 members, ${10 \choose 1} = 10$ have 1 member and ${10 \choose 2} = 45$ have 2 members. Thus the answer is $1024 - 1 - 10 - 45 = 968$.

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions