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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...

Circular Permutations

Circular Permutations are type of arrangements where the objects are to be arranged around a circle or in a circular order. Observe that in circular permutations the order around the circle (or the relative positions ) alone need to be taken into consideration and not the actual positions. For example, suppose $5$ different things are arranged around a circle. Consider the 5 positions around a circle. $A$, $B$, $C$, $D$, $E$ can be arranged in 5 different positions in 5! ways. Consider one such arrangement say $ABCDE$ in that order around the circle. This arrangement and the $4$ new arrangements $BCDEA$, $CDEAB$, $DEABC$, and  $EABCD$ are not only really different arrangements because the same relative positions around the circle are maintained by the $5$ letters. Therefore, the number of different ways of arranging the  $5$ letters around a circle is $ \cfrac {5!}{5} = 4!$ Also, in the above we are considering the clockwise and anti-clockwise arrangements on the circle ...

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 $4!$ arra...

Circular Permutations

In circular permutations, things are to be arranged in the form of a ring or a circle, e.g. arrangements of people around a circular table. In circular permutation there are no end points, i.e. there are no beginning or ending positions. So, the number of circular permutations of n objects are $$\frac{n!}{n}=(n-1)!$$ Thus in a circular permutation, one thing is kept fixed and the remaining  $(n-1)$ things are arranged in $(n-1)!$ ways. If the clockwise and counter clockwise orders are not distinguishable, then the number of ways = $\frac{1}{2}(n-1)!$ Let us look at a few examples.  Ex:   In how many ways can $6$ boys be seated at a circular table? Sol:   We keep one boy in a fixed position and to find the number of permutations to arrange the remaining 5 boys, we simply take $5!$, which gets us to our answer $\boxed {120} $.  Ex:  In how many ways can $6$ boys be arranged at a round table so that $2$ particular boys can be seat...

Permutations

Each of the different arrangements that can be made out of a given number of things by taking some or all of them at a time is called a permutation. Thus the permutation of the three letters $a, b, c$ taken two at a time are: $$ab, ba, bc, cb, ac, ca$$ Therefore the number of permutations of three different things taken two a time is $_{3}P_{2}$ or $P(3, 2) = 6$. We can generalize this to say $_{n}P_{r} =$ Total number of permutations of $n$ distinct things taken $r$ at a time. $$_{n}P_{r} = \frac{n!}{(n-r)!}$$ Below are some extensions of the above concept. Permutations of $n$ different things taken all at a time $$ = _{n}P_{n} = n!$$ Ex . Find the number of ways the letters of the word RAINBOW can be rearranged. Sol . Every letter in the word RAINBOW is different, so to get the number of permutations, we would simply use $n!$. $$n = 7 \Rightarrow n! = \boxed{5040}$$ Permutations of $n$ different things taken $r$ at a time, when $k$ things never occurs $$ = _{n-k}P_{...

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