Latest TCS Aptitude Question SOLUTION: There are 10 points on a straight line AB and 8 on another straight line AC none of them being point A. how many triangles can be formed with these points as vertices?

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There are 10 points on a straight line AB and 8 on another straight line AC
none of them being point A. how many triangles can be formed with these points
as vertices?

Asked In TCS Brindha D (6 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Sravya  (6 years ago)

10C2*8C1+8C2*10C1
=640

Is this solution Helpfull? Yes (14) | No (3)  

Krishna Vamsy  (6 years ago)

Ans: 720
You can choose any two points on AB and one point on AC : (10C2)*(8C1)
Similarly , you can choose any two points on AC and one point on AB: (8C2)*(10C1)
and Since two lines intersect at a you can join two points each on AB and AC with a line to form a triangle , therefore - (8 * 10)

so , (10C2)*(8C1) + (8C2)*(10C1) + 80 = 720

Is this solution Helpfull? Yes (8) | No   

sai  (6 years ago)

My answer assumes that you can't use A, B or C but only the other un-named points.

Use any 2 from the 10 on line AB with 1 of those from the 8 on AC = (10C2)*(8C1)

Use any 2 from the 8 on line AC with 1 of those from the 10 on AB = (8C2)*(10C1)

Is this solution Helpfull? Yes (3) | No   

nikita  (6 years ago)

10c2*8c1+8c2*10c1=360+280=640

Is this solution Helpfull? Yes (1) | No   

swarnalichakraborty  (6 years ago)

How?? Plz explain

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Gowthami  (6 years ago)

=10c2*8c1+10c1*8c2
=360+280
=540

Is this solution Helpfull? Yes | No (6)  

Divya Mathur  (6 years ago)

We have 19 points total. We need to choose 3 of them. So the formula is 19C3=969 triangles

Is this solution Helpfull? Yes | No (1)  

Pavani  (3 years ago)

Here given condition is A is excluded,so the formula to find the no.of triangles is
No.of triangles=mn(m+n)/2
m=10
n=8
No of triangles=10*8(10+8)/2
=180*9
= 720

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