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Answer these questions based on the following informations
There are n points on a plane, out of which m are collinear points, such that `mltn`
Find the maximum number of triangles formed by joining them.

A

`n!-m!`

B

`^mC_3-^nC_3`

C

`(n-m)C_3-^mC_3`

D

`^nC_3-^mC_3`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the maximum number of triangles that can be formed by joining n points on a plane, where m of those points are collinear, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Total Points**: We have a total of n points on the plane. Among these, m points are collinear. 2. **Triangles from n Points**: If there were no collinear points, the maximum number of triangles that could be formed by choosing any three points from these n points is given by the combination formula \( \binom{n}{3} \). This is calculated as: \[ \binom{n}{3} = \frac{n!}{3!(n-3)!} \] 3. **Triangles from Collinear Points**: However, since m points are collinear, any triangle formed by selecting three points from these m collinear points will not actually form a triangle. Thus, the number of ways to choose three points from these m collinear points is given by \( \binom{m}{3} \): \[ \binom{m}{3} = \frac{m!}{3!(m-3)!} \] 4. **Actual Triangles Formed**: To find the actual number of triangles that can be formed, we need to subtract the number of non-triangle combinations (from collinear points) from the total combinations of triangles (from all points). Therefore, the maximum number of triangles formed is: \[ \text{Maximum Triangles} = \binom{n}{3} - \binom{m}{3} \] 5. **Final Answer**: Thus, the answer to the question is: \[ \text{Maximum number of triangles} = \binom{n}{3} - \binom{m}{3} \] ### Conclusion: The final answer is \( \binom{n}{3} - \binom{m}{3} \).
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Knowledge Check

  • Answer these questions based on the following informations There are n points on a plane, out of which m are collinear points, such that mltn Find the maximum number of straight lines formed by joining them.

    A
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    B
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    A
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    B
    `^nC_4+^mC_4`
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    A
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    B
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