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26 Dec 2016, 11:12
Kudos
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D
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Difficulty:
5%
(low)
Question Stats:
82% (00:54) correct
18%
(01:08)
wrong
based on 192
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The circle above has 6 points on its edge. How many different triangles can be drawn that use three of these points as its vertices?
A. 2
B. 6
C. 15
D. 20
E. 30
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26 Dec 2016, 12:01
Kudos
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Bunuel
No three points are collinear, that is if we choose any 3 points on a given circle we can build a triangle.
Hence our problem boils down to choosing 3 points from the given six.
6C3 = 20
Answer D
Sirakri
GMAT 2: 700 Q48 V38
Posts: 104
27 Dec 2016, 10:33
Kudos
Bookmarks
Bunuel
Any 3 points (of the points forming a circle) can form a traingle.
Here we have specific 6 points => 6c3 (combinations because any order can be taken)
=> \(\frac{6!}{(6-3)!3!}\)
=> \(\frac{6*5*4*3!}{3!*(3*2)}\)
=> \(5*4\)
=> 20
D.
I've seen a similar question in which the question asks for the number of equilateral triangles :D
