Three points are marked on a line, and five points are marked on anoth

Question

Three points are marked on a line, and five points are marked on another, parallel to the first. The number of triangles that can be drawn using any 3 of these 8 points is:

(A) 90
(B) 45
(C) 42
(D) 26
(E) 25

ocelot22 · Answer

If you draw two lines in the xy plane, one with three points and one with five points, you should see that two points from one line will have to be mapped to one point on a second line to create a triangle. We have two choices. We can either map two points from the line with 5 points to one point from the line with three, or we can map in the reverse manner. See diagram

So the number of triangles = 5C2*3C1 + 3C2*5C1 = 30 +15= 45. OA is B

TarunKumar1234 · Answer

Three points are marked on a line, and five points are marked on another, parallel to the first. The number of triangles that can be drawn using any 3 of these 8 points is:

(A) 90
(B) 45
(C) 42
(D) 26
(E) 25

Show Answer

Official Answer

B

MathRevolution · Answer

To form a triangle we need 3 points out of which 2 points must be collinear (from the same line) and the remaining point should be from another line.

Hence,  ^3{C_2} * ^5{C_1} + ^5{C_2} * ^3{C_1}

=> 3 * 5 + 10 * 3 = 15 + 30 = 45

Answer B

MathRevolution · Answer

I believe there is a small error in calculation:

=>  ^5{C_2} = 10  and  ^3{C_1} = 3  so 10 * 3 = 30 and  ^3{C_2} = 3  and  ^5{C_1} = 5  so 3 * 5 = 15

=> Total: 30 + 15 = 45

Thanks

ScottTargetTestPrep · Answer

Solution:
 
Since we have two lines, and since a triangle has three points which are not collinear, two of the points must be on one line and the remaining point must be on the other line. Let line 1 be the line with 3 points and line 2 be the line with 5 points. We have two cases: 1) two points from line 1 and one point from line 2, and 2) one point from line 1 and two points from line 2.

Case 1: Two points from line 1 and one point from line 2

The number of triangles that can be formed in this scenario is 3C2 x 5C1 = 3 x 5 = 15.

Case 2: One point from line 1 and two points from line 2.

The number of triangles that can be formed in this scenario is 3C1 x 5C2 = 3 x 10 = 30.

Therefore, the total number of triangles that can be formed is 15 + 30 = 45.

Answer: B

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Three points are marked on a line, and five points are marked on anoth

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Three points are marked on a line, and five points are marked on anoth

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