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# A twelve-sided polygon consists of vertices A – L. How many

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Senior Manager
Joined: 20 May 2014
Posts: 282
Followers: 14

Kudos [?]: 49 [0], given: 220

A twelve-sided polygon consists of vertices A – L. How many [#permalink]  29 Oct 2017, 01:46
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Question Stats:

100% (01:32) correct 0% (00:00) wrong based on 1 sessions
A twelve-sided polygon consists of vertices A – L. How many lines can be drawn between any two vertices, such that a line is neither repeated, nor redundant, with any sides of the polygon?

(A) 45

(B) 54

(C) 55

(D) 66

(E) 110

Kudos for correct solution.
[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 327 [1] , given: 66

Re: A twelve-sided polygon consists of vertices A – L. How many [#permalink]  30 Oct 2017, 08:30
1
KUDOS
Let's start by computing the number of ways in which we can choose 2 points out of 12, i.e. $$12C2 = \frac{12!}{10!2!} = \frac{11*12}{2} = 66$$. Then, in order to eliminate the lines that joint two adjacent vertices, i.e. sides, we have to subtract 12. Thus, we get 66 - 12 = 54. Answer B
Re: A twelve-sided polygon consists of vertices A – L. How many   [#permalink] 30 Oct 2017, 08:30
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