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Number of Quadrilateral that can be made using the vertex of a polygon of $10$ sides as there vertices and having

(i) Exactly $1$ sides common with the polygon

(ii) Exactly $2$ sides common with the polygon

(iii) Exactly $3$ sides common with the polygon

juantheron
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2 Answers2

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A start: (i) The side common with a side of the polygon can be chosen in $10$ ways. Take one of these ways. In how many ways can we choose the remaining $2$ vertices? The neighbours of the vertices we already have can't be used. So we must choose $2$ points from the remaining $6$ so that these two points are not neighbours.

There are $\binom{6}{2}$ ways to choose $2$ points from $6$. But there are $5$ "bad" choices.

André Nicolas
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  • Thanks André Nicolas Got it for (ii) i have calculate using your method first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by $\binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \binom{5}{1} = 50$ but answer given is $=75$ I did not understand it. Thanks – juantheron Feb 07 '13 at 04:24
  • Two ways to get $2$ sides in common with polygon. Your type, two adjacent sides of the polygon, and a fourth point taken from $5$, so $50$. Also one edge of the polygon, and a non-adjacent other edge. Count these. There is a good chance you will get $50$ (wrong) because you may think of it as choosing one edge, then one edge. But that doble-counts these, so there are $25$ of this type. Finally, get your $50$ plus the $25$. – André Nicolas Feb 07 '13 at 05:23
  • What's a "bad" choice? I'm unable to understand what significance the bad choice has. Otherwise, I understand completely... – Abhigyan Nov 25 '18 at 03:33
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For exactly one common side 10×6c2= 150 For exactly 2 coomon side 10×5c2= 100 For exactly 3 common side 10×4c2 = 60

Prince jha
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