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How many triangles can be formed using 8 points in a given p [#permalink]
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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
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Re: Permutations 1 [#permalink]
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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]
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23 Aug 2009, 17:19
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The meaning of "8C31" 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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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]
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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]
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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]
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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]
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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 8C31=561=55 (Using rohansherry's method faster method)



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



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Re: Permutations 1 [#permalink]
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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]
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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]
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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]
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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]
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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



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Re: Problem: triangle in a plane Level: Medium How many [#permalink]
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10 Dec 2013, 05:20



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Re: How many triangles can be formed using 8 points in a given p [#permalink]
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23 Dec 2014, 08:57
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 noncollinear 8C3 no.s of distinct triangle can be formed. Hence, answer will be 8C31
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Re: How many triangles can be formed using 8 points in a given p [#permalink]
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29 May 2016, 05:15
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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19 Jun 2017, 13:32
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We have 8 points, and 3 are collinear:  Select 2 points among 5 noncollinear points (5C2) and 1 among the 3 collinear points ( x3 ): 3x5C2  Select 2 points among 3 collinear points (3C2) and 1 among 5 noncollinear ( x5 ) : 5x3C2  Select 3 points among the 5 noncollinear points: 5C3 Add all of them = 55 > equivalent to 8C3  1




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