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

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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 = \boxed{968}</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 = \boxed{968}</math>.
  
Note <math>{N \choose 0}+{N \choose 1} + {N \choose 2} + \dots + {N \choose N}</math> is equivalent to <math>2^N</math>
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Note <math>{n \choose 0}+{n \choose 1} + {n \choose 2} + \dots + {n \choose n}</math> is equivalent to <math>2^n</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 08:30, 27 October 2018

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 $1024 - 1 - 10 - 45 = \boxed{968}$.

Note ${n \choose 0}+{n \choose 1} + {n \choose 2} + \dots + {n \choose n}$ is equivalent to $2^n$

See also

1989 AIME (ProblemsAnswer KeyResources)
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

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