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Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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28 May 2012, 01:03

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macjas wrote:

Attachment:

Q19.gif

ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

A. 10 B. 15 C. 20 D. 25 E. 30

Generally in a plane if there are \(n\) points of which no three are collinear, then: 1. The number of triangles that can be formed by joining them is \(C^3_n\).

2. The number of quadrilaterals that can be formed by joining them is \(C^4_n\).

3. The number of polygons with \(k\) sides that can be formed by joining them is \(C^k_n\).

Since ABCDE is a regular pentagon then no three point out of 6 (5 vertices + center) will be collinear, so the number of triangles possible is \(C^3_6=20\).

Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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23 Sep 2013, 22:50

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Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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25 Sep 2015, 02:14

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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07 Oct 2015, 04:56

Bunuel wrote:

macjas wrote:

Attachment:

Q19.gif

ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

A. 10 B. 15 C. 20 D. 25 E. 30

Generally in a plane if there are \(n\) points of which no three are collinear, then: 1. The number of triangles that can be formed by joining them is \(C^3_n\).

2. The number of quadrilaterals that can be formed by joining them is \(C^4_n\).

3. The number of polygons with \(k\) sides that can be formed by joining them is \(C^k_n\).

Since ABCDE is a regular pentagon then no three point out of 6 (5 vertices + center) will be collinear, so the number of triangles possible is \(C^3_6=20\).

Answer: C.

Hope it helps.

Why the equation does not work for a rectangular consisting of ABCD and center F. I applied the equation 5C3=10 triangles while when I count them manually the result is 8 triangles. where is the mistake?

Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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07 Oct 2015, 05:41

Expert's post

1

This post was BOOKMARKED

Mo2men wrote:

Bunuel wrote:

macjas wrote:

Attachment:

Q19.gif

ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

A. 10 B. 15 C. 20 D. 25 E. 30

Generally in a plane if there are \(n\) points of which no three are collinear, then: 1. The number of triangles that can be formed by joining them is \(C^3_n\).

2. The number of quadrilaterals that can be formed by joining them is \(C^4_n\).

3. The number of polygons with \(k\) sides that can be formed by joining them is \(C^k_n\).

Since ABCDE is a regular pentagon then no three point out of 6 (5 vertices + center) will be collinear, so the number of triangles possible is \(C^3_6=20\).

Answer: C.

Hope it helps.

Why the equation does not work for a rectangular consisting of ABCD and center F. I applied the equation 5C3=10 triangles while when I count them manually the result is 8 triangles. where is the mistake?

Any help please

Please pay attention to the highlighted part.

In a rectangle the center is collinear with the endpoints of diagonals. _________________

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