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Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Correct answer is '160'. Can you explain this answer?
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Let AB, CD, EF, GH, and JK be five diameters of a circle with center ...
To find the number of ways to choose three points out of A, B, C, D, E, F, G, H, J, K, and O to form a triangle, we need to consider the different cases and combinations.
First, let's analyze the possible combinations of points that can form a triangle using the given diameters of the circle.
1. Choosing three points on the same diameter:
- There are 5 diameters in total (AB, CD, EF, GH, and JK).
- For each diameter, we can choose 3 points out of the 5 points on that diameter.
- So, the total number of triangles formed in this case = 5 * C(5, 3) = 5 * 10 = 50.
2. Choosing two points on one diameter and one point on another diameter:
- There are 5 diameters, and we need to choose 2 points from one diameter and 1 point from another diameter.
- The number of ways to choose 2 points from one diameter = C(5, 2) = 10.
- The number of ways to choose 1 point from another diameter = C(5, 1) = 5.
- So, the total number of triangles formed in this case = 5 * 10 * 5 = 250.
3. Choosing one point on each of three different diameters:
- There are 5 diameters, and we need to choose 1 point from each diameter.
- The number of ways to choose 1 point from each diameter = C(5, 1) * C(4, 1) * C(3, 1) = 5 * 4 * 3 = 60.
- However, in this case, we have counted each triangle 3 times (since there are 3 different orders to choose the points).
- So, the total number of triangles formed in this case = 60 / 3 = 20.
4. Choosing one point on one diameter, one point on another diameter, and one additional point:
- There are 5 diameters, and we need to choose 1 point from each of 2 diameters and 1 additional point.
- The number of ways to choose 1 point from each of 2 diameters = C(5, 1) * C(4, 1) = 5 * 4 = 20.
- The number of ways to choose 1 additional point = C(8, 1) = 8 (excluding the already chosen points and the center of the circle).
- So, the total number of triangles formed in this case = 20 * 20 * 8 = 3200.
Therefore, the total number of ways to choose three points to form a triangle = 50 + 250 + 20 + 3200 = 3520.
However, we need to consider the case where the three chosen points are collinear and do not form a triangle. In this case, the three points must be selected from the diameters AB, CD, EF, GH, or JK.
- Choosing three collinear points on the same diameter:
- There are 5 diameters in total (AB, CD, EF, GH, and JK).
- For each diameter, we can choose
First, let's analyze the possible combinations of points that can form a triangle using the given diameters of the circle.
1. Choosing three points on the same diameter:
- There are 5 diameters in total (AB, CD, EF, GH, and JK).
- For each diameter, we can choose 3 points out of the 5 points on that diameter.
- So, the total number of triangles formed in this case = 5 * C(5, 3) = 5 * 10 = 50.
2. Choosing two points on one diameter and one point on another diameter:
- There are 5 diameters, and we need to choose 2 points from one diameter and 1 point from another diameter.
- The number of ways to choose 2 points from one diameter = C(5, 2) = 10.
- The number of ways to choose 1 point from another diameter = C(5, 1) = 5.
- So, the total number of triangles formed in this case = 5 * 10 * 5 = 250.
3. Choosing one point on each of three different diameters:
- There are 5 diameters, and we need to choose 1 point from each diameter.
- The number of ways to choose 1 point from each diameter = C(5, 1) * C(4, 1) * C(3, 1) = 5 * 4 * 3 = 60.
- However, in this case, we have counted each triangle 3 times (since there are 3 different orders to choose the points).
- So, the total number of triangles formed in this case = 60 / 3 = 20.
4. Choosing one point on one diameter, one point on another diameter, and one additional point:
- There are 5 diameters, and we need to choose 1 point from each of 2 diameters and 1 additional point.
- The number of ways to choose 1 point from each of 2 diameters = C(5, 1) * C(4, 1) = 5 * 4 = 20.
- The number of ways to choose 1 additional point = C(8, 1) = 8 (excluding the already chosen points and the center of the circle).
- So, the total number of triangles formed in this case = 20 * 20 * 8 = 3200.
Therefore, the total number of ways to choose three points to form a triangle = 50 + 250 + 20 + 3200 = 3520.
However, we need to consider the case where the three chosen points are collinear and do not form a triangle. In this case, the three points must be selected from the diameters AB, CD, EF, GH, or JK.
- Choosing three collinear points on the same diameter:
- There are 5 diameters in total (AB, CD, EF, GH, and JK).
- For each diameter, we can choose
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Community Answer
Let AB, CD, EF, GH, and JK be five diameters of a circle with center ...
The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160
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Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer?
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Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer?.
Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?Correct answer is '160'. Can you explain this answer?.
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