Difference between revisions of "1989 AIME Problems/Problem 2"
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== Solution == | == 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 <math>2^{10} = 1024</math> total subsets of a ten-member [[set]], but of these <math> | + | 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 == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 18:15, 4 July 2013
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 
total subsets of a ten-member set, but of these 
have 0 members, 
have 1 member and 
have 2 members. Thus the answer is 
.
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
