Out of $18$ points in a plane, no three are in same straight line except $5$ points which are collinear.

Then the number of $(a)$ Straight lines $(b)$ Triangles which can be formed by joining them.

$\bf{My\; Solution::}$ $(a)$ For calculation of straight lines::

If there is no condition, then the number of straight line $ = $ choosing $2$ points out of $18$ points $\displaystyle = \binom{18}{2}$

but given that $5$ points are collinear. So we can draw only one line by joining them.

But I did not understand why my answer is wrong.

please explain me,



For straight lines, your strategy is correct. There are $\binom{18}{2}$ ways to choose $2$ points. But any choice of $2$ points from the $5$ collinear ones, which can be done in $\binom{5}{2}$ ways, produces the same line. Thus the total number of lines is $\binom{18}{2}-\binom{5}{2}+1$.

For triangles, there are $\binom{18}{3}$ ways to choose $3$ points, but $\binom{5}{3}$ of the choices are forbidden.

  • $\begingroup$ Nice solution André Nicolas. but i did not understand why we add 1 in last.Thanks $\endgroup$ – juantheron May 13 '14 at 4:49
  • $\begingroup$ You are welcome. When I subtracted $\binom{5}{2}$, I took away all lines formed by taking two points from the five. But there is one line. $\endgroup$ – André Nicolas May 13 '14 at 5:10

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.