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Snehaaaaa
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If one of two parallel lines has 7 points on it and the other has 8 25 Oct 2022, 01:22
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75% (01:43) correct 25% (01:51) wrong based on 453 sessions

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If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 455
B. 364
C. 196
D. 168
E. 56­


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If one of two parallel lines has 7 points on it and the other has 8 25 Oct 2022, 04:20
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Snehaaaaa
There are 7 points on a straight line, L1, and another 8 points on another line, L2 that is parallel to L1. How many triangles can be formed using these 15 points?

A. 455
B. 364
C. 196
D. 168
E. 56
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Two possible cases here
1) 2 points from 7 points line, and 1 from 8 points line
2) 1 point from 7 points line, and 2 from 8 points line

(7c2)*(8c1)+(8C2)*(7C1)
=168+196
=364
Option B
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If one of two parallel lines has 7 points on it and the other has 8 28 Apr 2024, 03:01
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I have a questionr regarding this one. Dont we need to divide 364 by 2 as a triangle ACB is the same as ABC etc.? Or am I missing something
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If one of two parallel lines has 7 points on it and the other has 8 28 Apr 2024, 05:55
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waddap
I have a questionr regarding this one. Dont we need to divide 364 by 2 as a triangle ACB is the same as ABC etc.? Or am I missing something
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Hope below would help.
­
If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 455
B. 364
C. 196
D. 168
E. 56


Approach #1:

There are two types of triangles that can be formed:

1. Triangles with two vertices on the line with 8 points and the third vertex on the line with 7 points: \(C^2_8*C^1_7=28*7=196\);

2. Triangles with two vertices on the line with 7 points and the third vertex on the line with 8 points: \(C^2_7*C^1_8=21*8=168\);

Total number of triangles: \(196+168=364\).

Approach #2:

Any three distinct points chosen from the total of \(8+7=15\) points will form a triangle, except for the cases where the three points are collinear.

Therefore, the total number of triangles can be calculated as \(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\). Here, \(C^3_8\) and \(C^3_7\) represent the number of different sets of 3 collinear points possible from the line with 8 points and the line with 7 points, respectively.


Answer: B
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If one of two parallel lines has 7 points on it and the other has 8 21 May 2025, 10:37
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Hi waddap, In this case we are selecting the points, rather than arranging them.

waddap
I have a questionr regarding this one. Dont we need to divide 364 by 2 as a triangle ACB is the same as ABC etc.? Or am I missing something
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