2
$\begingroup$

This is how I went about doing it. I know that if there are 'n' number of straight lines they intersect each other in nC2 ("n choose 2") ways. Therefore, the points of intersection using 6 lines is 15. I also know that a straight line and a circle intersect each other at atmost two different points. Hence the points of intersection of six lines and five circles is 6C1 x 5C1 x 2 = 60.

After this point I am a little confused. My answer was present in the options but it was the wrong one. I calculated it as being 15+60=75. But the answer key says it is 95. Their reason being that 2 circles intersect in atmost two different points and therefore the points of intersection of 5 circles is 5C2 x 2 = 20. The answer then works out to 15+60+20 = 95.

I did not understand the last part. Could someone please help me clear out my confusion? Thanks.

$\endgroup$
6
$\begingroup$

There are exactly $3$ types of intersection points:

  1. Between a line and a line (${}_6C_2 \cdot 1=15$).
  2. Between a line and a circle (${}_6C_1 \cdot {}_5C_1 \cdot 2=60$).
  3. Between a circle and a circle. This is the case that you were forgetting. There are $5$ circles in total, so there are ${}_5C_2$ ways to choose the two circles that will intersect. These two circles can intersect at most $2$ times. This yields ${}_5C_2 \cdot 2 = 20$.

Summing everything together, we obtain $15 + 60 + 20 = 95$.

$\endgroup$
2
  • $\begingroup$ Thanks. But it would be 5C1 in your point 2 right? Or am I understanding wrong? $\endgroup$ – Sharon Jul 10 '13 at 5:47
  • $\begingroup$ Whoops, that was a typo. You are correct. $\endgroup$ – Adriano Jul 10 '13 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.