In a plane there are 16 non collinear points find the number of straight lines formed

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There are $25$ points on a plane of which $7$ are collinear . How many quadrilaterals can be formed from these points ?

I did this $^{25}C_{4}-^{7}C_{4}=12615$ quadrilaterals.

But the book is giving answer $^{25}C_{4}-^{7}C_{4}-^{7}C_{3}\times ^{18}C_{1}=11985$ quadrilaterals..

I don't know what is the correct idea .

I look for a short and simple way.

I have studied maths up to $12$th grade.

asked Sep 4, 2015 at 1:11

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Your idea is very close to correct. A quadrilateral is formed by $4$ points, where at most $2$ may be colinear. Thus, we have $$ \underbrace{^{25} C_4}_{\text{ choose four points}} - \underbrace{^{7} C_4}_{\text{ subtract out ways to pick four colinear points}} - \underbrace{^{7} C_3 \cdot ^{18} C_1 }_{\text{ subtract out ways to pick three colinear points}}$$

answered Sep 4, 2015 at 1:24

Marcus MMarcus M

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History of Indian Constitution

15 Questions 15 Marks 9 Mins

Concept:

To Find the Straight lines formed on 8 non collinear points

We know that the straight line has two points.

That is, it has starting point and ending points

∴ According to permutation and combination we have

Number of lines = 8c2

Expand the above combination value.

⇒ (8×7)/(2×1)

⇒ 4× 7

⇒ 28 lines

∴ 28 straight lines are formed from 8 non - collinear points on the X-Y plane.

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Answer

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Hint: We will use the fact that 2 points form a line and then find the lines formed at max by these 10 points. After that for the b part, we will just subtract the lines which are to be ruled out because of 4 points being in a straight line and thus the answer.

Complete step by step answer:
We know that we need at least 2 points (distinct) to form a straight line. So, the maximum number of lines formed by these 10 points will be that number of lines where no 3 points lie in a single line.
Part a):
We need to choose 2 points among the 10.
Hence, the number of lines formed by 10 points, no three of which are collinear will be $^{10}{C_2}$.
We know that $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
Hence, $^{10}{C_2} = \dfrac{{10!}}{{2!(10 - 2)!}}$
On simplifying it, we will get:-
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10!}}{{2! \times 8!}}$
This is equivalent to:-
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}}$
On simplifying it further, we will get as follows:-
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10 \times 9}}{{2 \times 1}} = 45$.

Hence, the answer of part a) is 45 lines.

Part b):
We basically here lost 4 of the points which originally were involved in the previous part.
So, let us find out how many lines these 4 points were forming earlier.
These 4 points will contribute to forming $^4{C_2}$.
We know that $^4{C_2} = \dfrac{{4!}}{{2!(4 - 2)!}}$.
Hence, $^4{C_2} = \dfrac{{4!}}{{2! \times 2!}}$
This is equivalent to:-
${ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3 \times 2!}}{{2 \times 1 \times 2!}}$
On simplifying it further, we will get as follows:-
${ \Rightarrow ^4}{C_2} = 2 \times 3 = 6$.

Hence, we will get 45 – 6 + 1 lines that are 40 lines.

Note:
The students might be confused about the reason why we added 1 after subtracting 6 from 45. But the students must note that these 4 lines are actually forming one single line. When we subtracted the lines forming from these 4 points, we excluded that as well. Therefore, we added 1 back again.
Additional Information:- Permutation and Combination have made our life extremely easy and relaxed. We do not actually have to find all the combinations possible but we can directly find the number possible.

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How many straight lines can be drawn from non collinear points?

How many straight lines can be formed from 8 non collinear points on the XY plane A 19859 B 18 C 28 D 56?

What is the formula of non collinear points?