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How many triangles can be formed using 8 points in a given p

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How many triangles can be formed using 8 points in a given p [#permalink] New post 23 Aug 2009, 04:46
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How many triangles can be formed using 8 points in a given plane?

(1) A triangle is formed by joining 3 distinct points in the plane
(2) Out of 8 given points, three are collinear
[Reveal] Spoiler: OA
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 05:08
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IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 07:55
rohansherry wrote:
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.


Good one! kudos goes to you.
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 11:17
flyingbunny wrote:
rohansherry wrote:
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.


Good one! kudos goes to you.



thanks ya... was wondering when will i get my first kudos
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 17:19
The meaning of "8C3-1" is really brief, clear and precise. That is the beauty of math. I like it and you deserve a kudos for this. :)
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 20:09
rohansherry wrote:
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.


Agree with B. But I think the combination formula might be wrong.

IMHO, only 3 points are collinear, all the other 5 points are not on the same line. Hence, each of the 3 points can be combined with 2 of the 5 points to create a triangle.

Therefore, the # of triangles = 3 x 5C2 = 3 x 10 = 30.

Last edited by Jivana on 23 Aug 2009, 20:22, edited 1 time in total.
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 20:20
Jivana wrote:
rohansherry wrote:
IMO B. 8c3 - 1

A is the general stmt. we know that Triangle is fromed using only 3 points.


Agree with B. But I think the combination formula might be wrong.

IMHO, only 3 points are collinear, all the other 5 points are not on the same line. Hence, each of the 3 points can be combined with 2 of the 5 points to create a triangle.

Therefore, the # of triangles = 3 x 5C3 = 3 x 10 = 30.


You forgot 2 of the 3 points (collinear) and 1 of the 5 points to form a triangle.
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 20:26
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.
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Re: Permutations 1 [#permalink] New post 23 Aug 2009, 23:32
Jivana wrote:
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.



I just feel you are missing here something.... Can yo explain how you comingup with this "3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45"
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Re: Permutations 1 [#permalink] New post 24 Aug 2009, 06:48
rohansherry wrote:
Jivana wrote:
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.



I just feel you are missing here something.... Can yo explain how you comingup with this "3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45"

Yup. It is not necessary to select ATLEAST one point from the three collinear points (as per the above equation)
SO we have to add 5C3 = 10 too.

So answer is 45+10 = 55 (Using Jivana's conventional method) and 8C3-1=56-1=55 (Using rohansherry's method faster method) :)
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Re: Permutations 1 [#permalink] New post 24 Aug 2009, 06:57
Economist wrote:
rohansherry wrote:
Jivana wrote:
Yep, I did indeed forget that.

# should be: 3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45.



I just feel you are missing here something.... Can yo explain how you comingup with this "3 x 5C3 + 5 x 3C2 = 3 x 10 + 5 x 3 = 45"

Yup. It is not necessary to select ATLEAST one point from the three collinear points (as per the above equation)
SO we have to add 5C3 = 10 too.

So answer is 45+10 = 55 (Using Jivana's conventional method) and 8C3-1=56-1=55 (Using rohansherry's method faster method) :)


I believe it is the fastest. :-D
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Re: Permutations 1 [#permalink] New post 24 Aug 2009, 12:43
Agreed, 8C3 - 1 is the best method.
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Re: Permutations 1 [#permalink] New post 25 Aug 2009, 05:35
HI Guys, My answer would also be B.
But, I am not clear in the possibilities to form triangles.
1. using the 3 non collinear : 5C3
2. using one collinear and 2 non collinear : 3C1x5C2

I am sure that I am missing something here.
Would you please be of any help to understand the 55 possibilities?

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Re: Permutations 1 [#permalink] New post 25 Aug 2009, 05:39
Guys,

I just noticed that I was missing the 2 of the 3 points (collinear) and 1 of the 5 points

Thx
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Re: Permutations 1 [#permalink] New post 25 Aug 2009, 05:41
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

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Re: Permutations 1 [#permalink] New post 25 Aug 2009, 06:16
defoue wrote:
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds


the triangle by three points that are collinear.
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Re: Permutations 1 [#permalink] New post 25 Aug 2009, 06:19
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defoue wrote:
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds


hey see,

if have to make all triangles from n points...then it would be nc3.....because triangle has 3 sides.... so out of n any 3 sides...

now 3 point are colinear so we cant make one triangle...so we se subtract that from it..


hope its clear....now
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Re: Permutations 1 [#permalink] New post 26 Aug 2009, 03:59
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rohansherry wrote:
defoue wrote:
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

Rgds


hey see,

if have to make all triangles from n points...then it would be nc3.....because triangle has 3 sides.... so out of n any 3 sides...

now 3 point are colinear so we cant make one triangle...so we se subtract that from it..


hope its clear....now


Cristal clear
Thx very much
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Re: Permutations 1 [#permalink] New post 26 Aug 2009, 04:27
Guys, I am sorry that it took me so long to post the OA. But here it is:

1. is insufficient because it just states a well known fact
2. is sufficient because in such case we can calculate the number of triangles that can be formed: 5C3+8*3C2+3C1*5C2

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Re: Problem: triangle in a plane Level: Medium How many [#permalink] New post 10 Dec 2013, 05:09
Why can't we just pick all possible combinations of 3 points out of 8 to answer the first question?
8C3 looks sufficient to me
please help
Re: Problem: triangle in a plane Level: Medium How many   [#permalink] 10 Dec 2013, 05:09

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