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Senior Manager Joined: 20 May 2014
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How many unique quadrilaterals can be inscribed in the verti [#permalink] 00:00

Question Stats: 0% (00:00) correct 100% (01:54) wrong based on 1 sessions
How many unique quadrilaterals can be inscribed in the vertices of a nonagon (a 9-sided figure), if points A and B, two vertices in the nonagon, cannot make up the same quadrilateral?

(A) 126

(B) 105

(C) 96

(D) 65

(E) 21

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[Reveal] Spoiler: OA Director Joined: 03 Sep 2017
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Re: How many unique quadrilaterals can be inscribed in the verti [#permalink]
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The total number of quadrilaterals that can be inscribed in a nonagon can be computed by counting the number of ways 4 points can be chosen out of 9, i.e. $$9C4 = \frac{9!}{5!4!} = 126$$.

But this is not what we are looking for. Indeed, we need to exclude those quadrilaterals which have two points being A and B. To compute them, given that two points are fixed, we can count how many ways are there to choose 2 points out of 7, i.e. $$7C2 = \frac{7!}{2!5!} = 21$$. Subtracting 21 from 126, we get 105. Target Test Prep Representative Status: Head GRE Instructor
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Re: How many unique quadrilaterals can be inscribed in the verti [#permalink]
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Expert's post
Bunuel wrote:
How many unique quadrilaterals can be inscribed in the vertices of a nonagon (a 9-sided figure), if points A and B, two vertices in the nonagon, cannot make up the same quadrilateral?

(A) 126

(B) 105

(C) 96

(D) 65

(E) 21

Since a nonagon has 9 vertices and a quadrilateral has 4 vertices, the number of quadrilaterals that can be made is 9C4 = (9 x 8 x 7 x 6)/4! = (9 x 8 x 7 x 6)/(4 x 3 x 2) = 3 x 7 x 6 = 126, if there are no restrictions. However, since vertices A and B can’t both be in the same quadrilateral, we need to subtract the number of quadrilaterals that have both vertices A and B. The number of such quadrilaterals is 2C2 x 7C2 = 1 x (7 x 6)/2! = 42/2 = 21 (notice that 2C2 is the number of ways A and B can be picked if they have to be 2 vertices of the quadrilateral and 7C2 is the number of ways the other 2 vertices of the quadrilateral can be picked from the other 7 vertices).

Thus, the number of quadrilaterals such that vertices A and B are not in the same quadrilateral is 126 - 21 = 105.

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500+ lessons 3000+ practice problems 800+ HD solutions Re: How many unique quadrilaterals can be inscribed in the verti   [#permalink] 19 Dec 2017, 06:32
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