Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Permutations - When some objects are Not Distinct

Let us consider the following example - In how many different ways can the letters of the word 'POSSESSIVE' be arranged?

 Observe that there are $10$ letters in the word and all these letters are not distinct. There are $4$ $S's$ and $2E's$ and other letters are distinct. If all the letters were distinct, the number of different arrangements is obviously $^{10}P_{10}$ or $10!$. In this case, since some of the letters are alike the number of arrangements cannot be $10!$ 

 Let us assume that the number of arrangements of the letters of the word 'POSSESSIVE' is $x$. Consider any of these permutations (or arrangements) A typical arrangement can be thought as PSOESVSIES. Keep all except the three $S's$ fixed in their places. 

 Suppose we replace the four $S's$ in the above by four distinct objects, say $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}.$ If we permute these objects among themselves in all possible ways, it will give rise to $^{4}P_{4}$  or $4!$ arrangements. Therefore, by changing the $4S's$ into $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}.$ (i.e. by changing these like things to different or distinct things) and by permuting these $4$ things among themselves the $x$ arrangements become $x \cdot 4!.$

 Now, consider one among the above $ x \cdot 4! $ arrangements. If similarly we replace the two $E's$ by $E_{1}$ and $E_{2}$ and permute these two among themselves, the total number of permutations becomes $x \cdot 4! \cdot 2!$. But this is the number of permutations of $10$ different letters taken all at a time, which we know is $10!$. Therefore, 

 $x \cdot 4! \cdot 2! = 10!$  $ \Rightarrow $ $x = \cfrac {10!}{4!2!}$

 This above example leads us to the following result relating to the number of permutations when some of the things or objects are distinct (or are alike)

 Result: 

The number of permutations (or arrangements) of $n$ things taken all at a time, when the $n$ things are made up of $p$ like things of one sort, $q$ like things for another sort, $r$ like things of another sort and so on, is given by $ \cfrac {n!}{[p!q!r!...]}$

 For example, the number of $6$ digit numbers that can be formed by using all the digits $3, 3, 6, 6, 6, 8$ is $ \cfrac {6!}{3!2!} = 60$. In this, the $6$ digits are not distinct, $6$ is occurs 3 times, and $3$ is occurs twice. 

 Your turn - 

 $(i)$ How many different numbers can be formed with the digits $4, 4, 4, 4, 5, 5, 1, 3, 3, 6, 7, 3$ using all the digits each time?

 $(ii)$ How many of these are divisible by 5?

 Leave a comment for solutions or additional practice problems. 

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Equilateral and Equiangular Polygons

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