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Bunuel
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Bunuel
How many different quadrilaterals can be formed if lines are drawn connecting the vertices of a regular pentagon?

A. 1
B. 4
C. 5
D. 6
E. 10

Since a pentagon has 5 points on its edge and we need to create quadrilaterals, each of which has 4 vertices, the number of ways to create those quadrilaterals using the 5 points is:

5C4 = 5!/4!1! = 5

Answer: C
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I was tricked by the wording of this question. I understood the question as : lines are drawn from the vertices, then we count the number of quadrilaterals after that. This would include quadrilaterals that have each vertex as one of the vertex of the original pentagon vertices, AND quadrilaterals that have some vertices the intersection between the newly drawn lines.
Hence , to me the answer was : 5C4(for the first case, quads drawn with each vertex as vertex of the pentagon) + additional quads with atleast one vertex not part of the original pentagon vertex.

Any thoughts about the wording of the problem?
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no of diagonal formula = n(n-3)/2
where n is the number of sides
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