Latest Infosys Aptitude Question SOLUTION: There are 10 points in three parallel planes, first plane contain 5 points, second plane contain 3 points and third plan contain rest of them, then how many traingles to be formed

• Placement Papers
• /
• Infosys
• /
• Category

Infosys Question Papers: Infosys Interview Experience(10) INFOSYS Company Profile

(#M40165364) INFOSYS Category question puzzle Keep an EYE

There are 10 points in three parallel planes, first plane contain 5 points, second plane contain 3 points and third plan contain rest of them, then how many traingles to be formed ?

1)104
2)109
3)105
4)106
5)None of these

Asked In Infosys manoj (5 years ago)

Unsolved Read Solution (9)

Is this Puzzle helpful?   (62)   (22) Submit Your Solution

Read Solution (9) :

shaik imran  (5 years ago)

For forming a triangle we need 3 points therefore from 10 points we can form 10c3 triangles but
out of 10 points 5(1st line),3(2nd line),2(3rd line) points are collinear so we have to eliminate those points so that sub those colinear points from 10c3
ie, =(10c3)-(5c3)-(3c3)
=120-10-1
=109

Is this solution Helpfull? Yes (18) | No (2)  

Rex  (5 years ago)

For forming a triangle,
1. You need 3 points
2. Those points should not be in the same plane.

Let us name the planes as p2, p3, p5 where the the number denotes the number of points it has
Therefore if you calculate every possible case:
consider taking 2 points from 1st plane and then 1 point from 2nd plane that is
1. Take 2 points from p2 and one from p3 i.e = 2c2 x 3c1 = 3
2. Take 2 points from p2 and one from p5 i.e = 2c2 x 5c1 = 5
3. Take 2 points from p3 and one from p5 i.e = 3c2 x 5c1 = 15
4. Take 2 points from p3 and one from p2 i.e = 3c2 x 2c1 = 6
5. Take 2 points from p5 and one from p2 i.e = 5c2 x 2c1 = 20
6. Take 2 points from p5 and one from p3 i.e = 5c2 x 3c1 = 30

Now one more case remain in which you can choose 1 point from each plane = 5c1 x 3c1 x 2c1 = 30

Adding all up (30 + 30 +20 + 15 + 5 + 6 + 3) we get 109 [Answer]

Is this solution Helpfull? Yes (6) | No   

FAISAL  (5 years ago)

Lets consider:
Plane 1: 2 points
Plane 2 : 3 points
Plane 3: 5 points

No. of triangle b/w Plane 1 & 2: 4C3 + 4C3
No. of triangles b/w Plane 1 &3: 6C3 + 6C3
No. of triangles b/w Plane 2&3 : 6C3 + 6C3 +6C3

Summation of all these we get: 108

Is this solution Helpfull? Yes (3) | No (16)  

Thanga Melki  (5 years ago)

109
Line 1 - 5 points can make 4 combination of each 5*4 = 20(each point in plane 2 with plane 1) so 20*3 - 60
Same follow for plane 1to 3 so 20*2 - 40.
For plane 2 to 3 - 9
SO 60+40+9=109
We can say--5^3 - 75
5^2-25
3^2 - 9
75+25+9 = 109

Is this solution Helpfull? Yes (3) | No (1)  

PRADEEP KUMAR  (5 years ago)

3 point needs to form triangle.

5 point are collinear,in one plane and 3 point are collinear in other plane.these point will not form triangle.
so 10c3-5c3-3c3=109 triangles

Is this solution Helpfull? Yes (2) | No   

Soma  (5 years ago)

The Tips & Tricks to solve this problem is Here https://youtu.be/NwOOffNdNjM

Is this solution Helpfull? Yes | No (6)  

Bhupesh Mishra  (5 years ago)

109 10c3-5c3-3c3 =109

Is this solution Helpfull? Yes | No   

harsh kumar  (5 years ago)

10c3-5c3-3c3=109

Is this solution Helpfull? Yes | No   

Chandramohan  (3 years ago)

The answer is 109

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.