
There are $n$ points in a plane of which p points are collinear. How many lines can be formed from these points?
1) ${ }^{(n-p)} C_{2}$
2) ${ }^{n} C_{2}-{ }^{p} C_{2}$
3) ${ }^{n} C_{2}-{ }^{p} C_{2}+1$
4) ${ }^{n} C_{2}-{ }^{P} C_{2}-1$
Answer
216k+ views
Hint: Collinear points are those that are situated along a single or shared straight line. Collinear points are those that are on a line either close to or far from another point or more than two points.
Formula Used:
${ }^{n} C_{r}=\dfrac{n !}{r ! .(n-r) !}$
Complete step by step Solution:
Combinations are ways to choose things from a collection where the order of the choices is irrelevant. Let's say we have a trio of numbers: P, Q, and R. Then, the combination determines how many ways we can choose two numbers from each group.
It is easy to count the number of combinations in smaller cases, but the likelihood of a set of combinations is also higher in cases with many groups of components or sets.
Given that there are n points in the plane and that p of those points are collinear,
$t \mathrm{~s}={ }^{\mathrm{n}} \mathrm{C}_{2}$ is the number of lines created from n points.
Lines of numbers produced from p collinear points $={ }^{\mathrm{P}} \mathrm{C}_{2}$
The number of lines with all p points is equal to 1.
Thus, the total number of lines generated is equal to $={ }^{\mathrm{n}} \mathrm{C}_{2}-{ }^{\mathrm{P}} \mathrm{C}_{2}+1$
Therefore, the correct option is 3.
Note: Utilizing the combinations formula, it is simple to determine how many distinct groups of r objects each may be created from the provided n unique objects. The factorial of n divided by the product of the factorial of r and the factorial of the difference between n and r, respectively, is the formula for combinations.
${ }^{n} C_{r}=\dfrac{n !}{r ! .(n-r) !}$
Formula Used:
${ }^{n} C_{r}=\dfrac{n !}{r ! .(n-r) !}$
Complete step by step Solution:
Combinations are ways to choose things from a collection where the order of the choices is irrelevant. Let's say we have a trio of numbers: P, Q, and R. Then, the combination determines how many ways we can choose two numbers from each group.
It is easy to count the number of combinations in smaller cases, but the likelihood of a set of combinations is also higher in cases with many groups of components or sets.
Given that there are n points in the plane and that p of those points are collinear,
$t \mathrm{~s}={ }^{\mathrm{n}} \mathrm{C}_{2}$ is the number of lines created from n points.
Lines of numbers produced from p collinear points $={ }^{\mathrm{P}} \mathrm{C}_{2}$
The number of lines with all p points is equal to 1.
Thus, the total number of lines generated is equal to $={ }^{\mathrm{n}} \mathrm{C}_{2}-{ }^{\mathrm{P}} \mathrm{C}_{2}+1$
Therefore, the correct option is 3.
Note: Utilizing the combinations formula, it is simple to determine how many distinct groups of r objects each may be created from the provided n unique objects. The factorial of n divided by the product of the factorial of r and the factorial of the difference between n and r, respectively, is the formula for combinations.
${ }^{n} C_{r}=\dfrac{n !}{r ! .(n-r) !}$
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