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Permutation and Combination
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?
Option
a. 680
b. 720
c. 816
d. 640
Read Solution (Total 16)
-
- to form a triangle we need 3 points
select 2 points from the 10 points of line AB & 1 from the 8 on AC
= (10C2)*(8C1)
select 2 points from the 8 points of line AC & 1 from the 10 on AB=
(8C2)*(10C1)
total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
d.640 - 10 years agoHelpfull: Yes(97) No(5)
- 80 triangles also possible by taking point A as 3rd vertex
so ans is 640+80=720
- 10 years agoHelpfull: Yes(11) No(15)
- To form a triangle we need three point...here we have total 18 points.
So number of triangle will b 816.(18c3)
But 10 on same line and other 8 are also on same line eventually they cant form triangles..
So total number of triangle will b 18c3 -10c3 - 8c3 which is 816 -120 - 56 = 640 . - 10 years agoHelpfull: Yes(6) No(1)
- Ans: (10 c1 * 8c1)+ (10 c2 *8 c 1) +(8 c 2* 10 c 1)=720
- 8 years agoHelpfull: Yes(5) No(0)
- Ans is 640 as (10C2)*(8C1)+ (8C2)*(10C1) = 640
A triangle can be formed using 3 points so we can either chose 2 points from AB and 1 point from AC(10C2*8C1) or 1 point from AB and 2 points from AC(10C1*8C2).
- 10 years agoHelpfull: Yes(3) No(3)
- to form a triangle we need 3 points
select 2 points from the 10 points of line AC & 1 of from the 8 on AB
= (10C2)*(8C1)
select 2 points from the 8 points of line AB & 1 of from the 10 on AC=
(8C2)*(10C1)
total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
d.640 - 10 years agoHelpfull: Yes(2) No(4)
- to form a triangle we need 3 points
select 2 points from the 10 points of line AB & 1 from the 8 on AC
= (10C2)*(8C1)
select 2 points from the 8 points of line AC & 1 from the 10 on AB=
(8C2)*(10C1)
total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
d.640 - 10 years agoHelpfull: Yes(1) No(0)
- with vertice A there are 80 tringles possible, which are taken in to count with 640
- 9 years agoHelpfull: Yes(1) No(0)
- To form a triangle we need 3 vertices.So we can select one vertice from first straight line and two vertices from second straight line.In the same way we can have one more combination as shown below:
(10c2*8c1)+(10c1*8c2)+(10c1*8c1)=720.
The last combination (10c1*8c1) forms a triangle by taking single point from each straight line and thitd point as 'A'. - 7 years agoHelpfull: Yes(1) No(0)
- RAKESH can u plz explain how are u getting 640? I mean the calculation.
- 10 years agoHelpfull: Yes(0) No(0)
- to form a triangle we need 3 points
select 2 points from the 10 points of line AB & 1 from the 8 on AC
= (10C2)*(8C1)
select 2 points from the 8 points of line AC & 1 from the 10 on AB=
(8C2)*(10C1)
total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
d.640 - 10 years agoHelpfull: Yes(0) No(1)
- 10c2*8c1+10c1*8c2 =640
as 3 points are required to form a triangle - 10 years agoHelpfull: Yes(0) No(2)
- Atriangle can be formed by three points, this three point can be choose in this way
10c1*8c2 +10c*8c1= 640 - 9 years agoHelpfull: Yes(0) No(0)
- 10C2*8C1+10C1*8C2=640
- 5 years agoHelpfull: Yes(0) No(0)
- 10!*9!*8!+8!*7!*9! = 640
- 2 years agoHelpfull: Yes(0) No(0)
- To form a traingle we need 3 points . Let two points from straight line with 10 points and one point from another 8 point line..and vice versa by applying combination.. 10C2*8C1 + 8C2*10C1
- 1 year agoHelpfull: Yes(0) No(0)
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