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Question: Find the number of straight lines obtained in joining 10 points on a plane if a) No three of which...

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.

Explanation

Solution

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 10C2^{10}{C_2}.
We know that nCr=n!r!(nr)!^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}.
Hence, 10C2=10!2!(102)!^{10}{C_2} = \dfrac{{10!}}{{2!(10 - 2)!}}
On simplifying it, we will get:-
10C2=10!2!×8!{ \Rightarrow ^{10}}{C_2} = \dfrac{{10!}}{{2! \times 8!}}
This is equivalent to:-
10C2=10×9×8!2×1×8!{ \Rightarrow ^{10}}{C_2} = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}}
On simplifying it further, we will get as follows:-
10C2=10×92×1=45{ \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 4C2^4{C_2}.
We know that 4C2=4!2!(42)!^4{C_2} = \dfrac{{4!}}{{2!(4 - 2)!}}.
Hence, 4C2=4!2!×2!^4{C_2} = \dfrac{{4!}}{{2! \times 2!}}
This is equivalent to:-
4C2=4×3×2!2×1×2!{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3 \times 2!}}{{2 \times 1 \times 2!}}
On simplifying it further, we will get as follows:-
4C2=2×3=6{ \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.