Question

The maximum number of points of intersection of five lines and four circles is
(A). 60
(B). 72
(C). 62
(D). None of these

Answer 1

Hint: We’ll solve this question by counting the intersection points. We’ll use the combination method for counting. Also, intersection points are the points which satisfy both the curves. Here, we’ll see that maximum number of points, which can satisfy five lines and four circles.

Complete step-by-step answer:
First, we’ll observe the intersection points of two straight lines, two circles and one straight line and one circle.

Here we can easily see that, we can get one point from the intersection of two lines.
Now, we’ll see the intersection point of two circles.

We can clearly see that we’ll get 2 intersection points from the intersection of two circles. Now, we’ll see the intersection point of a line with a circle.

We can observe from the above figure that we’ll get 2 points from the intersection of a circle and line.
Now, let’s count the intersection points.
For straight lines, there are $${}^5{C}_2$$ ways to select 2 straight lines and from each pair of straight lines we’ll get 1 intersection point. So, the total number of intersection point will be $${}^5{C}_2\times 1={}^5{C}_2$$
 For circles, we have ${}^4{C}_2$ ways to choose two circles and from each pair of circles we’ll get 2 intersection points. So, the total number of points will be ${}^4{C}_2\times 2$
For one circle and One straight line, there are ${}^4{C}_1\times {}^5{C}_1$ ways to choose and with each pair we’ll get 2 intersection points. So, the total number of intersection points will be ${}^4{C}_1\times {}^5{C}_1\times 2$
Now, we have encountered all the intersection points. We’ll just calculate them by using the combination formula which is ${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$.
$${}^5{C}_2={\displaystyle \frac{5!}{2!3!}}={\displaystyle \frac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}}=10$$
${}^4{C}_2\times 2={\displaystyle \frac{4!}{2!2!}}\times 2={\displaystyle \frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}\times 2=12$
And, ${}^4{C}_1\times {}^5{C}_1\times 2={\displaystyle \frac{4!}{1!3!}}\times {\displaystyle \frac{5!}{1!4!}}\times 2={\displaystyle \frac{4\times 3!}{3!}}\times {\displaystyle \frac{5\times 4!}{4!}}\times 2=4\times 5\times 2=40$
So, the total number of intersection points will be 10+12+40=62
Hence Option C is the correct option.

Note:Those questions are really good in mathematics in which you can’t find the topic. It’s the matter of pure application. Usually, in questions of permutation and combination, students forget to take some cases and end-up with the wrong answer. Combination tell us the number of ways by which we can choose something,