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

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

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23 Sep 2003, 15:04
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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?

10
15
20
25
30
Intern
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Re: PS : Regular Pentagon [#permalink]

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23 Sep 2003, 15:34
Choose 3 points at a time out of 6 points:

6C3 = 20
CEO
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Re: PS : Regular Pentagon [#permalink]

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23 Sep 2003, 17:00
edealfan wrote:
Choose 3 points at a time out of 6 points:

6C3 = 20

. B

. A .F .C

.E .D

I got 20 , but i subtracted ABC,AED,BCD,BAE..etc..since they are already connected..

So finally i ended up with 10

I might be totally wrong here, can you clarify?

thanks
praetorian
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Re: PS : Regular Pentagon [#permalink]

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23 Sep 2003, 21:10
Not sure why you subtracted ABC,AED,BCD,BAE..etc. I interpreted the question stem as "maximum" possible triangles that can be formed by the points, A, B, C, D, E and F. We know they are all non-collinear. So, you can choose any three points and form a triangle.

Hope this helps.
Senior Manager
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24 Sep 2003, 00:49
thats correct. In such questions very important to note whether points are collinear or not.
Praet, try to solve this and u will know when to subtract sides:
How many triangles can be formed by connecting vertices of a hexagon such that no side of triangle coincides with that of hexagon.
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Re: PS : Regular Pentagon [#permalink]

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24 Sep 2003, 01:22
edealfan wrote:
Not sure why you subtracted ABC,AED,BCD,BAE..etc. I interpreted the question stem as "maximum" possible triangles that can be formed by the points, A, B, C, D, E and F. We know they are all non-collinear. So, you can choose any three points and form a triangle.

Hope this helps.

absolutely agree. the paints are not collinear and the question does not asks to count different triangles, so its solution should be simple 3C6=20.
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16 Jan 2004, 14:10
I did it in a slightly different way.

Chose a point and then chose 2 out of 5 remaining points
so we have 5C2 per point. For 6 points we have
6 * 5C2 triangles.
Since triangle ABC is same as BCA and CAB we need to divide the combinations by 3
so total triangles = 6 * 5C2 /3 = 20
16 Jan 2004, 14:10
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