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Numerical Ability
Age Problem
AB is a straight line and 10 points are drawn in it. AC another straight 8 points are drwan in it. how many triangle can be made with these points as a vertices?
Read Solution (Total 15)
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- 8c2*10c1 +8c1*10c2=640 ways
- 10 years agoHelpfull: Yes(20) No(1)
- 9c2*7c1 + 7C2*9c1 +7C1*9C1,,according to me i thnk we hv to consider triangles formd by common point A
- 10 years agoHelpfull: Yes(8) No(1)
- In AB line,10 points and in AC 8 points..
So,to form a triangle we need 3 points..
Thus,We can do it in (10c2*8c1)*(8c2*10c1) ways.... - 10 years agoHelpfull: Yes(5) No(8)
- We can think in two ways.
1. 18c3-10c3-8c3=640
2. 10c2*8c1 + 8c2*10c1=640 - 10 years agoHelpfull: Yes(5) No(0)
- Sorry,my given answer is wrong..It will be (10c2*8c1)+(8c2*10c1)..Sorry for the mistake ...
- 10 years agoHelpfull: Yes(1) No(1)
- 8c2*10c1 +8c1*10c2=640 ways
- 10 years agoHelpfull: Yes(1) No(0)
- 9c1*7c2 + 7c1*9c2 + 9c1*7c1
- 10 years agoHelpfull: Yes(1) No(0)
- I think A point will be common to both straight lines. So i think "9c1*7c1 + 10c2*7c1 + 10c1*7c2 + 8c1*9c2 + 8c2*9c1 = 1128 ways " will be the ans ...
- 10 years agoHelpfull: Yes(1) No(0)
- ans is 539
- 10 years agoHelpfull: Yes(0) No(3)
- sry it will be 594
- 10 years agoHelpfull: Yes(0) No(3)
- its 7c1*10c2+8c2*9c1
- 10 years agoHelpfull: Yes(0) No(0)
- 8c2*10c1 +8c1*10c2-7(because a first point a and 7 point of ac can n't make triangle with a)-9( because a first point a and 9 point of ab can n't make triangle with a)=624
- 10 years agoHelpfull: Yes(0) No(1)
- 624
8c2*10c1
- 10 years agoHelpfull: Yes(0) No(0)
- ANS: 640 ways
To form a triangle, we need 3 points
there are 2 ways of it:
1 is to take 2 points from points on AB and 1 point each from AC OR vice verse
= (10C2*8C1)+(10C1*8C2)= 640 WAYS - 10 years agoHelpfull: Yes(0) No(0)
- 1 point is common to both..
so ans will be
10c2*7c1+7c2*10c1+7c1*9c1 - 10 years agoHelpfull: Yes(0) No(0)
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