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How many unique quadrilaterals can be inscribed in the verti
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12 Nov 2017, 01:18
12 Nov 2017, 01:18
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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
Kudos for correct solution.
(A) 126
(B) 105
(C) 96
(D) 65
(E) 21
Kudos for correct solution.
Re: How many unique quadrilaterals can be inscribed in the verti
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13 Nov 2017, 00:24
13 Nov 2017, 00:24
1
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.
Answer B
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.
Answer B
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Re: How many unique quadrilaterals can be inscribed in the verti
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19 Dec 2017, 06:32
19 Dec 2017, 06:32
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Expert Reply
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
(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.
Answer: B
