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There are 10 points in a plane, no three of which are in the same straight line, except 4 points which are collinear. The total number of triangles that can be formed with the vertices as these points is:
- a)120
- b)124
- c)116
- d)112
Correct answer is option 'C'. Can you explain this answer?
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There are 10 points in a plane, no three of which are in the same str...
Problem Analysis:
We have 10 points in a plane, out of which 4 points are collinear and no three points are collinear. We need to find the total number of triangles that can be formed using these points as vertices.
Solution:
Step 1: Counting the number of ways to form a triangle
To form a triangle, we need to choose 3 points out of the given 10 points. The number of ways to choose 3 points out of 10 is denoted by "10C3" and can be calculated using the formula:
10C3 = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120
Step 2: Subtracting the number of triangles with collinear points
Out of the 10 points, 4 points are collinear. Let's assume these collinear points are A, B, C, and D. Any triangle formed using these collinear points will be a degenerate triangle, i.e., all three vertices will be on the same line. We need to exclude these degenerate triangles from our count.
Case 1: All three points are collinear
In this case, we can choose any 3 points out of the given 4 collinear points. The number of ways to choose 3 points out of 4 is denoted by "4C3" and can be calculated using the formula:
4C3 = 4! / (3! * (4-3)!) = 4! / (3! * 1!) = 4
Case 2: Two points are collinear
In this case, we can choose 2 points from the collinear points and 1 point from the remaining 6 non-collinear points. The number of ways to choose 2 points out of 4 collinear points is denoted by "4C2" and the number of ways to choose 1 point out of 6 non-collinear points is denoted by "6C1". The total number of triangles formed in this case is given by:
4C2 * 6C1 = (4! / (2! * (4-2)!)) * (6! / (1! * (6-1)!)) = (4! / (2! * 2!)) * (6! / (1! * 5!)) = (4 * 3 / 2 * 1) * 6 = 6 * 6 = 36
Step 3: Subtracting the degenerate triangles from the total count
The total number of triangles that can be formed with the given points is the number of ways to choose 3 points minus the number of degenerate triangles:
Total number of triangles = 10C3 - (4C3 + 4C2 * 6C1) = 120 - (4 + 36) = 120 - 40 = 80
Therefore, the correct answer is option C) 116.
We have 10 points in a plane, out of which 4 points are collinear and no three points are collinear. We need to find the total number of triangles that can be formed using these points as vertices.
Solution:
Step 1: Counting the number of ways to form a triangle
To form a triangle, we need to choose 3 points out of the given 10 points. The number of ways to choose 3 points out of 10 is denoted by "10C3" and can be calculated using the formula:
10C3 = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120
Step 2: Subtracting the number of triangles with collinear points
Out of the 10 points, 4 points are collinear. Let's assume these collinear points are A, B, C, and D. Any triangle formed using these collinear points will be a degenerate triangle, i.e., all three vertices will be on the same line. We need to exclude these degenerate triangles from our count.
Case 1: All three points are collinear
In this case, we can choose any 3 points out of the given 4 collinear points. The number of ways to choose 3 points out of 4 is denoted by "4C3" and can be calculated using the formula:
4C3 = 4! / (3! * (4-3)!) = 4! / (3! * 1!) = 4
Case 2: Two points are collinear
In this case, we can choose 2 points from the collinear points and 1 point from the remaining 6 non-collinear points. The number of ways to choose 2 points out of 4 collinear points is denoted by "4C2" and the number of ways to choose 1 point out of 6 non-collinear points is denoted by "6C1". The total number of triangles formed in this case is given by:
4C2 * 6C1 = (4! / (2! * (4-2)!)) * (6! / (1! * (6-1)!)) = (4! / (2! * 2!)) * (6! / (1! * 5!)) = (4 * 3 / 2 * 1) * 6 = 6 * 6 = 36
Step 3: Subtracting the degenerate triangles from the total count
The total number of triangles that can be formed with the given points is the number of ways to choose 3 points minus the number of degenerate triangles:
Total number of triangles = 10C3 - (4C3 + 4C2 * 6C1) = 120 - (4 + 36) = 120 - 40 = 80
Therefore, the correct answer is option C) 116.
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Community Answer
There are 10 points in a plane, no three of which are in the same str...
Number of triangles formed joining the 10 points taken 3 at a time = 10C3 = 120
Number of triangles formed joining the 4 points taken 3 at a time = 4C3 = 4
But four collinear points cannot form a triangle when taken 3 at a time
So the total number of required triangle = 120 - 4 = 116.
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There are 10 points in a plane, no three of which are in the same straight line, except 4 points which are collinear. The total number of triangles that can be formed with the vertices as these points is:a)120b)124c)116d)112Correct answer is option 'C'. Can you explain this answer?
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