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Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
How many Quadrilaterals can be formed from 10 points out of which 6
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06 Mar 2021, 04:36
06 Mar 2021, 04:36
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Re: How many Quadrilaterals can be formed from 10 points out of which 6
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06 Mar 2021, 05:29
06 Mar 2021, 05:29
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given
10 points -- 6 are colinear (set A)and 4 are non colinear (set B)
for quadrilateral we need 4 points
no of quadrilaterals by selecting 2 points from set A and 2 points from set B = (6C2) * (4C2) =90
no of quadrilaterals by selecting 4 points set B =(4C4) =1
no of quadrilaterals by selecting 1 points from set A and 3 points from set B = (6C1) * (4C3) =24
total no of quadrilaterals will be = 90+24+1 =115
10 points -- 6 are colinear (set A)and 4 are non colinear (set B)
for quadrilateral we need 4 points
no of quadrilaterals by selecting 2 points from set A and 2 points from set B = (6C2) * (4C2) =90
no of quadrilaterals by selecting 4 points set B =(4C4) =1
no of quadrilaterals by selecting 1 points from set A and 3 points from set B = (6C1) * (4C3) =24
total no of quadrilaterals will be = 90+24+1 =115
General Discussion
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: How many Quadrilaterals can be formed from 10 points out of which 6
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06 Mar 2021, 23:51
06 Mar 2021, 23:51
1
OA Explanation
Another method:
Notice we cannot make Quadrilaterals from those 6 points (Because it will be a Line) and we cannot pick 3 points from these 6 (because it will be a Triangle)
Number of ways in which we can pick 4 points out of 10 points = \(^{10}C_4 = 210\)
Number of ways in which we can pick 4 points out of those 6 points = \(^6C_4 = 15\)
Number of ways in which we can pick 3 points out of those 6 points and 1 point from other 4 points = \(^6C_3\) x \(^4C_1\) \(= (20)(4) = 80\)
Therefore, Number of Quadrilaterals \(= 210 - 15 - 80 = 115\)
Another method:
Notice we cannot make Quadrilaterals from those 6 points (Because it will be a Line) and we cannot pick 3 points from these 6 (because it will be a Triangle)
Number of ways in which we can pick 4 points out of 10 points = \(^{10}C_4 = 210\)
Number of ways in which we can pick 4 points out of those 6 points = \(^6C_4 = 15\)
Number of ways in which we can pick 3 points out of those 6 points and 1 point from other 4 points = \(^6C_3\) x \(^4C_1\) \(= (20)(4) = 80\)
Therefore, Number of Quadrilaterals \(= 210 - 15 - 80 = 115\)
