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Another Permutation and Combination question (Discussion)

Sanchit kshirsagar saidSat, 20 Dec 2008 03:23:02 -0000 ( Link )

There are 25 points on a plane which 7 are collinear. now solve the following?

How many quadrilateras can be formed from these points?

a)5206

B)2603

3)13015

4) None of these

I wish to know if there can be a formula that will help me solve the question. Like for triangles N(N-1)(N-2)/6

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  1. Sureshbala saidMon, 05 Jan 2009 20:13:29 -0000 ( Link )

    Dear Sanchit,

    We need not remember a formula to answer questions like these. Instead let us get into some logic.

    Let us say there are ‘n’ points on a plane of which no three are collinear.

    In order to draw a line since we need 2 points and from the given ‘n’ points we can select 2 points in ^nC_2 ways.

    Hence the total number of lines that can be drawn using these ‘n’ points is  ^nC_2

    Similarly for a triangle since we need 3 points, the total number of triangles that can be drawn will be ^nC_3 = \frac{n(n-1)(n-2)}{6}


    Now out of the given ‘n’ points if ‘m’ points are collinear,

    The total number of straight lines will be ^nC_2 - ^mC_2 + 1 (Here 1 is added because all the ‘m’ collinear points will give one line)

    The total number of triangles will be  ^nC_3 - ^mC_3

    So coming to your question the answer will be ^{25}C_4 - ^7C_4 =12615 .

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