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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,

Thanks

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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.

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  • $\begingroup$ 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

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