Find the number of straight lines obtained in joining 10 points on a plane if
a) No three of which are collinear
b) Four points are collinear.
Answer
649.2k+ views
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.
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.
Recently Updated Pages
Difference Between Prokaryotic Cells and Eukaryotic Cells

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

