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# ABCDE is a regular pentagon with F at its center. How many

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

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27 May 2012, 09:43
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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
[Reveal] Spoiler: OA
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Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]

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27 May 2012, 09:50
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Choosing any 3 points out of the 6 available points will form a triangle. Hence this turns into a combination question.

Choose 3 points out of 6: 6c3=20
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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 $$C^3_6=20$$.

Similar questions to practice:
m03-71107.html
the-sides-bc-ca-ab-of-triangle-abc-have-3-4-5-interior-109690.html
if-4-points-are-indicated-on-a-line-and-5-points-are-132677.html

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

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10 Jun 2012, 01:44
some body please clear this discrepancy

from 6 if we choose 3 , how do we denote it , all books and materials show this as 6C3

but here parrot man shows this as 6c3 and moderator shows this as 3c6 , leading me to becoming confused !!

according to me , the larger one goes at the top and the smaller sub group goes below

so from 6 if we choose 3 then 6C3.

so please if someone could confirm why is 3C6 denoted here , and does 3C6 mean the same as 6C3 ? are they same ?

so from n if if have to choose , 3 or 4 or 5 etc shouldn't it be nc3 , nc4 , nc5 etc .... Please clear my confusion
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Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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10 Jun 2012, 03:09
There are many forms of denoting the combinations.

nCk or C(n,k) or Ckn as written by bunuel, all are different standards.

But the meaning is same. Choosing 'k' objects from a group of 'n' objects.
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Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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23 Sep 2013, 22:50
Hello from the GMAT Club BumpBot!

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

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03 Jan 2014, 17:47
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

Combination Q; hw many ways to select 3 points from a group of 6 points . 6C3 = 6*5*4*3! / 3!*6 = 20

OA=C

Thanks for posting

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

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06 Sep 2014, 22:10
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
..

choosing 3points from 6 points.. 6c3= 20
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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).

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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 $$C^3_6=20$$.

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?

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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 $$C^3_6=20$$.

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?

Please pay attention to the highlighted part.

In a rectangle the center is collinear with the endpoints of diagonals.
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Re: ABCDE is a regular pentagon with F at its center. How many   [#permalink] 07 Oct 2015, 05:41
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