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Combinations

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by swerve » Fri Feb 26, 2021 10:49 am

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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

The OA is B

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Re: Combinations

by Scott@TargetTestPrep » Sat Mar 13, 2021 3:08 pm
swerve wrote: ↑
Fri Feb 26, 2021 10:49 am
 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

The OA is B

Solution:

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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