
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
584.7k+ 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.
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