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If there are p,q, r points on three parallel lines ,the number of triangles which can be formed with vertices at their points is
(p+q+r)C3− (pC3 + qC3 + rC3)
now, to get max no of triangles (pC3 + qC3 + rC3) will be minimum.
Here, p+q+r = 19
now, 19 can be distributed in p,q,r in several ways, such as (10,5,4) , (5,6,8), (6,6,7) , ... and so on. from max- min concept when all the values are distributed equally , the resultant value will give either max or min. so here the logical combination will be (6,6,7)

so, the max required triangles can be formed 19C3- 6C3-6C3-7C3 = 969-20-20-35 = 894.
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another approach, please correct if wrong..

Question summary: 19 points plotted 3 parallel lines, how many triangles can be made?

=> 19 points and to be distributed in 3 groups.
=> 19C3 = 969

Now for 19 points in 3 lines, each line A, B, C should have max 6,6,7 points for equal distribution since we need maximum lines available for cross intersections (triangle formation)

Therefore number of ways 3 points can be selected from each line A, B, C are 6C3 , 6C3, 7C3
Sum = 6C3 + 6C3 + 7C3 = 75 (this defys triangle property since no 3 points can be on same line)

These ways should be subtracted from the total to get number of triangles available = 969 - 75 = 894
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Let's assume three parallel lines A, B, C where A has 7 and B and C have 6 points each.

First scenario:

Taking 2 points from A and 1 point from B to plot triangles.

Number of Triangles: 7C2 X 6C5 = 126.

Similarly, taking 2 points from A and 1 point from C will give us 126 more triangles. (126+126 = 252 triangles so far)

Second scenario:

Taking 1 point from A and 2 points from B to plot triangles.

Number of Triangles: 7C1 X 6C2 = 105.

Similarly, taking 1 point from A and 2 points from C will give us 105 more triangles. (105+105+252 = 462 triangles so far)

Third scenario:

Taking 2 points from B and 1 point from C to plot triangles.

Number of Triangles: 6C4 X 6C5 = 90.

Similarly, taking 1 point from B and 2 points from C will give us 90 more triangles. (90+90+462 = 642 triangles)

Total triangles: 642 (Option: D)
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hi, official OA states the answer to be 642, while most of people are getting 894.
can someone please clarify what is the correct answer?
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Dillesh4096
What is the maximum number of triangles that can be formed by plotting 19 points on three parallel lines?

A. 300
B. 416
C. 540
D. 642
E. 894
Solution:

Assuming no three points are collinear, if each point is picked from a different parallel line, then the maximum number of triangles that can be formed occurs when the number of points on each parallel line is the same or almost the same. Since there are 19 points, we can have 6 points of each of the two of the parallel lines and 7 points on the last parallel lines (notice that 19/3 = 6.33).

Triangles can be formed in the following ways:

1) Each vertex of the triangle is picked from a distinct parallel line. The maximum number of triangles that can be formed is 6 x 6 x 7 = 252 (assuming no three points are collinear).

2) Two vertices of the triangle are picked from one parallel line and the third vertex is picked from another parallel line. The number of triangles that can be formed is (6C2 x 6 + 6C2 x 7 + 7C2 x 6) x 2 = 321 x 2 = 642.

Therefore, the maximum number of triangles that can be formed is 252 + 642 = 894.

Answer: E
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Dillesh4096
What is the maximum number of triangles that can be formed by plotting 19 points on three parallel lines?

A. 300
B. 416
C. 540
D. 642
E. 894
Solution:

Assuming no three points are collinear, if each point is picked from a different parallel line, then the maximum number of triangles that can be formed occurs when the number of points on each parallel line is the same or almost the same. Since there are 19 points, we can have 6 points of each of the two of the parallel lines and 7 points on the last parallel lines (notice that 19/3 = 6.33).

Triangles can be formed in the following ways:

1) Each vertex of the triangle is picked from a distinct parallel line. The maximum number of triangles that can be formed is 6 x 6 x 7 = 252 (assuming no three points are collinear).

2) Two vertices of the triangle are picked from one parallel line and the third vertex is picked from another parallel line. The number of triangles that can be formed is (6C2 x 6 + 6C2 x 7 + 7C2 x 6) x 2 = 321 x 2 = 642.

Therefore, the maximum number of triangles that can be formed is 252 + 642 = 894.

Answer: E

Thanks for the explanation !!how do we assume the points are not collinear? should not we substract the case when they are collinear?
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