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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]

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New post 23 Aug 2009, 05:46
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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
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Re: Permutations 1 [#permalink]

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New post 23 Aug 2009, 06:08
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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]

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New post 23 Aug 2009, 18:19
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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]

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New post 23 Aug 2009, 21:09
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, 21:22, edited 1 time in total.

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Re: Permutations 1 [#permalink]

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New post 23 Aug 2009, 21:20
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]

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New post 23 Aug 2009, 21:26
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]

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New post 24 Aug 2009, 00:32
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]

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New post 24 Aug 2009, 07:48
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]

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New post 25 Aug 2009, 06:35
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]

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New post 25 Aug 2009, 06:39
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]

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New post 25 Aug 2009, 06:41
Can someone please explain the "8C3 - 1" method.
What does this "-1" stands for?

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Re: Permutations 1 [#permalink]

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New post 25 Aug 2009, 07:16
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]

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New post 25 Aug 2009, 07:19
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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]

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New post 26 Aug 2009, 04:59
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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]

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New post 26 Aug 2009, 05:27
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]

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New post 10 Dec 2013, 06:09
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

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Re: Problem: triangle in a plane Level: Medium How many [#permalink]

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Re: How many triangles can be formed using 8 points in a given p [#permalink]

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New post 23 Dec 2014, 09:57
Question doesnt mention the orientation of 8 points. What if all points are collinear.
So, statement 1) is ruled out
Statement 2) Three points are collinear, they cannot form a triangle. By combination if all points are non-collinear 8C3 no.s of distinct triangle can be formed.
Hence, answer will be 8C3-1
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Re: How many triangles can be formed using 8 points in a given p [#permalink]

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New post 29 May 2016, 06:15
CasperMonday wrote:
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


The answer here is not b but E.

Comment: point (x,y) and points (a,b) are identical if [x=a] and [y=b].

1. A is a general stetment, and in addition, we canno't form a trianle in which 2 or more points are identicle.
-> since we do not know how many points are identical:
-> not suff.
2. Exacly 3 points are co linear, this say nothing about how many points are identical.
2 points can be coliner but identical.
-> not suff

Hence, Choice E.

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Re: How many triangles can be formed using 8 points in a given p [#permalink]

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New post 19 Jun 2017, 14:32
We have 8 points, and 3 are collinear:
- Select 2 points among 5 non-collinear points (5C2) and 1 among the 3 collinear points ( x3 ): 3x5C2
- Select 2 points among 3 collinear points (3C2) and 1 among 5 non-collinear ( x5 ) : 5x3C2
- Select 3 points among the 5 non-collinear points: 5C3
Add all of them = 55 --> equivalent to 8C3 - 1

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Re: How many triangles can be formed using 8 points in a given p   [#permalink] 19 Jun 2017, 14:32

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