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.


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 $15 + 60 + 20 = 95$.

  • $\begingroup$ 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

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