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Question: In a plane there are \[37\] straight lines, of which \[13\] pass through the point A and \[11\] pass...

In a plane there are 3737 straight lines, of which 1313 pass through the point A and 1111 pass through the point B. Besides, no three lines pass through one point, no line passes through both point A and B, and no two are parallel. Then the number of intersection points the lines have is equal to-
(A) 535535
(B) 601601
(C) 728728
(D) None of these

Explanation

Solution

In this question, we need to find out the number of intersection points from the given particular.
A point of intersection means the point in which two lines intersect. We need to consider all the given particulars and find out the total possible point of intersection. Then subtracting the points, we have calculated extra we will get the required result.
Formula to be used:
C(n,r)=nCr=n!r!(n−r)!C(n,r){ = ^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}

Complete answer:
We have, in the plane there are 3737 straight lines.
In which 1313 pass through point A and 1111 pass-through point B. Also, no three lines pass through one point, no line passes through both A and B and no two are parallel.
We need to find out the number of intersection points the lines have.
A point of intersection means the point in which two lines intersect.
In the plane, there are 3737 straight lines that have 37C2^{37}{C_2} points of intersection since we need two straight lines to create a point of intersection and the total numbers are thirty-seven.
1313 straight lines pass through point A yield only one intersection point instead of 13C2^{13}{C_2} and 1111 straight lines pass through point B also yield only one intersection point instead 11C2^{11}{C_2}.
Therefore, the number of total intersection points =37C2−13C2−11C2+1+1^{37}{C_2}{ - ^{13}}{C_2}{ - ^{11}}{C_2} + 1 + 1
=37!2!(37−2)!−13!2!(13−2)!−11!2!(11−2)!+1+1\dfrac{{37!}}{{2!(37 - 2)!}} - \dfrac{{13!}}{{2!(13 - 2)!}} - \dfrac{{11!}}{{2!(11 - 2)!}} + 1 + 1
=37!2!35!−13!2!11!−11!2!9!+2= \dfrac{{37!}}{{2!35!}} - \dfrac{{13!}}{{2!11!}} - \dfrac{{11!}}{{2!9!}} + 2
=37×36×35!2×1×35!−13×12×11!2×1×11!−11×10×9!2×1×9!+2= \dfrac{{37 \times 36 \times 35!}}{{2 \times 1 \times 35!}} - \dfrac{{13 \times 12 \times 11!}}{{2 \times 1 \times 11!}} - \dfrac{{11 \times 10 \times 9!}}{{2 \times 1 \times 9!}} + 2
=37×18−13×6−11×5+2= 37 \times 18 - 13 \times 6 - 11 \times 5 + 2
=666−78−55+2= 666 - 78 - 55 + 2
=535= 535
Hence, the number of intersection points equals 535535.

Therefore, option (A) is the correct option.

Note:
Combination:
The combination is represented as a selection of items from a collection, such that the order of selection does not matter.
For a combination,
C(n,r)=nCr=n!r!(n−r)!C(n,r){ = ^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
Where, the factorial of n is denoted by n! and defined by,
n!=n(n−1)(n−2)....2×1n! = n(n - 1)(n - 2)....2 \times 1