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Bunuel
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How many unique quadrilaterals can be inscribed in the vertices of a 19 Jul 2017, 23:24
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B
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59% (02:09) correct 41% (02:05) wrong based on 378 sessions

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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
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B
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How many unique quadrilaterals can be inscribed in the vertices of a 25 Jul 2017, 11:24
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Bunuel
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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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
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mcmoorthy
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How many unique quadrilaterals can be inscribed in the vertices of a Updated on: 20 Jul 2017, 03:56
Originally posted by mcmoorthy on 20 Jul 2017, 01:56.
Last edited by mcmoorthy on 20 Jul 2017, 03:56, edited 1 time in total.
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IMHO B

Total number of quadrilaterals that can be formed from a nonagon is 9C4 = 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is 2C2 * 7C2 = 21

Now number of quadrilaterals with A and B not together is 126 - 21= 105

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How many unique quadrilaterals can be inscribed in the vertices of a 20 Jul 2017, 03:39
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mcmoorthy
IMHO B

Total number of quadrilaterals that can be formed from a nonagon is 9C4 = 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is 2C2 * 8C2 = 21

Now number of quadrilaterals with A and B not together is 126 - 21= 105

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Hi Murthy,
The first part of your explanation I understand, but the second part is not clear. Can you please help with that .


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How many unique quadrilaterals can be inscribed in the vertices of a 20 Jul 2017, 03:48
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mcmoorthy
IMHO B

Total number of quadrilaterals that can be formed from a nonagon is 9C4 = 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is 2C2 * 8C2 = 21

Now number of quadrilaterals with A and B not together is 126 - 21= 105

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I think you've made a small mistake over there, it should be 7c2 since you have already selected 2 points and you will have to select other 2 points from remaining 7 points.

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How many unique quadrilaterals can be inscribed in the vertices of a 20 Jul 2017, 03:57
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Edited. I had written 8 instead of 7. The number is 21 anyway. Thanks for the nudge.

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How many unique quadrilaterals can be inscribed in the vertices of a 26 Jul 2021, 02:50
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Total number of quadrilaterals that can be formed from a nonagon is 9C4
=>choding 4 point out of 9 therefore
=> 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is
2C2 * 7C2 = 21

Therefore the total no of points that can be formed is 126 - 21= 105
Hence IMO B
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How many unique quadrilaterals can be inscribed in the vertices of a 31 Jul 2021, 00:44
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Hi anyone,

Don't we have to chose 2 vertices A and B.... or is it like, it is already given?
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How many unique quadrilaterals can be inscribed in the vertices of a 07 Apr 2025, 04:35
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Are we sure about this question...

It asks for "unique" quadras and if you mark each vertices from 1-9 or A-I (for ease of illustration but each of the vertices are essentially identical if you rotate the diagram), assume the nonagon is a regular nonagon where all sides are congruent in length, and all angles are equal in measure, quadra A-C-D-E would be the same as A-H-G-F, etc. due to symetry. There aren't as many as 105.
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