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

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

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27 May 2012, 08:43
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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:
Attachment:

Q19.gif [ 3.29 KiB | Viewed 10565 times ]
[Reveal] Spoiler: OA
Intern
Joined: 10 Apr 2011
Posts: 14
Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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27 May 2012, 08:50
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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
Math Expert
Joined: 02 Sep 2009
Posts: 43896
ABCDE is a regular pentagon with F at its center. How many [#permalink]

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28 May 2012, 00:03
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macjas wrote:

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:
http://gmatclub.com/forum/m03-71107.html
http://gmatclub.com/forum/the-sides-bc- ... 09690.html
http://gmatclub.com/forum/if-4-points-a ... 32677.html

Hope it helps.
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Posts: 450
Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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10 Jun 2012, 00:44
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, 02:09
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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Joined: 26 Mar 2013
Posts: 1442
Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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07 Oct 2015, 03:56
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?

Math Expert
Joined: 02 Sep 2009
Posts: 43896
Re: ABCDE is a regular pentagon with F at its center. How many [#permalink]

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07 Oct 2015, 04:41
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]

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05 Apr 2017, 00:02
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Re: ABCDE is a regular pentagon with F at its center. How many   [#permalink] 05 Apr 2017, 00:02
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