Of the 12 points in a given plane 5 are on circle O and remaining 7 ar

Question

Of the 12 points in a given plane 5 are on circle O and remaining 7 are on line L. Number of different triangles that can be formed by these 12 points are 
(Assume: 2 points on a circle and 1 point on the line are not collinear)

A. 70
B. 175
C. 185
D. 276
E. 462

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kt9696 · Accepted Answer

Of the 12 points in a given plane 5 are on circle O and remaining 7 are on line L. Number of different triangles that can be formed by these 12 points are 
(Assume: 2 points on a circle and 1 point on the line are not collinear)

A. 70
B. 175
C. 185
D. 276
E. 462

Here's my approach:

Total points in the given plane = 12
Total triangles possible (if none of them were collinear) = 12C3
BUT, 7 points are collinear, so 7C3 triangles cannot be formed.

So, 12C3 - 7C3 = 185 triangles

nick1816 · Answer

Bunuel  Shouldn't this mention in the question ( 2 points on a circle and 1 point on the line are not collinear)

Total number of triangles
= 5C3+7C1*5C2+7C2*5C1
=10+70+105=185

Archit3110 · Answer

we are considering what is the max. no of ∆ that can be formed using all the points
5c3+ 5c1*7c2 + 7c1*5c2 = 105+ 70 ; 185
IMO C

exc4libur · Answer

cases circle + line:
5C2*7C1=70
5C1*7C2=105

cases circle:
5C3=10

total=175+10=185

Ans (C)

jfranciscocuencag · Answer

Can someone plz give an explanation?

Kind regards!

shiva1325 · Answer

as given 
there are 5 points on circle O and 7 points on line L
now case 1: where triangles are formed by the points taking only on circle
This is done in 5c3 ways=10
case 2: where 2 points on circle and one point on Line (as no two points on cirlce and one point on straight line are collinear)
this can be done in 5c2*7c1 ways=70
case 3: where 1 point on circle and 2 points on straight line
this can be done in 5c1*7c2 ways=105
=>10+70+105=185 ways a triangle can be formed

Fdambro294 · Answer

3 scenarios (since only the 7 points on the line will be the only time when 3 or more points are collinear)

Scenario 1:
Choose any 2 points from the circle and 1 point from the line:

5! / (2! 3!) * 7 = 70 ways

OR

Scenario 2: choose any 2 points from the line and a 3rd point from the circle

7! / (2! 5!) * 5 = 105

OR

Scenario 3: can create a triangle completely inscribed in the circle by choosing any 3 points from the circle’s circumference

5! / (3! 2!) = 10

70 + 105 + 10 =

185

Posted from my mobile device

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Of the 12 points in a given plane 5 are on circle O and remaining 7 ar

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Of the 12 points in a given plane 5 are on circle O and remaining 7 ar

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