Lesson Permutations

Source code of 'Permutations'

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By the permutations of the letters abc we mean all of their possible arrangements:
<blockquote>abc
acb
bac
bca
cab
cba</blockquote>

There are 6 permutations for three things.  As the number of things (letters in this case) increases, their permutations grow 
astronomically; for example, if twelve different things are permuted, then the number of their permutations is 479,001,600.

This number was derived theoretically from the FUNDAMENTAL PRINCIPLE OF COUNTING. 
<blockquote> If something can be chosen, or can happen, or be done, in m different ways, and, after that has happened, something else
can be chosen in n different ways, then the number of ways of choosing both of them is {{{m*n}}}.</blockquote>

For example, imagine the letters a, b, c, d put into a hat, and let us draw two of them in succession. We can draw the first in 4
different ways: either a or b or c or d. After that has happened, there are 3 ways to choose the second. Therefore, there are
{{{4*3}}} or 12 possible ways to choose two letters from four.
<blockquote>ab
ba
ca
da 
ac
bc
cb
db
ad
bd
cd
dc</blockquote>

b means that a was chosen first and b second; ba means that b was chosen first and a second; and so on.

Let us now consider the total number of permutations of all four letters. There are 4 ways to choose the first. 3 ways remain
to choose the second, 2 ways to choose the third, and 1 way to choose the last. Therefore the number of permutations of 4 different things is
<blockquote>4� 3� 2� 1 = 24.</blockquote>

That is, the number of permutations of 4 different things taken 4 at a time  is 4!.

In general, the number of permutations of n different things taken n at a time is n!.

EXAMPLE 1

Five different books are on a shelf.  In how many different ways could you arrange them?

Answer:{{{5!=5*4*3*2*1=120}}}.

EXAMPLE 2

There are 6! permutations of the 6 letters of the word square.
a)  In how many of them is r the second letter?  _ r _ _ _ _
b)  In how many of them are q and e next to each other?

Answer:a) Let r be the second letter. Then there are 5 ways to fill the first spot, 4 ways to fill the third, 3 to fill the fourth,
and so on. There are 5! such permutations. {{{5!=120}}}.

b)  Let q and e be next to each other as qe.  Then we will be permuting the 5 units qe, s, u a, r.  They have 5! permutations.  But q and e
could be together as eq.  Therefore, the total number of ways they can be next to each other is 2� 5! = 240.

We have seen that the number of ways of choosing 2 letters from 4 is 4� 3 = 12. This is called
<blockquote>The number of permutations of 4
different things taken 2 at a time.</blockquote>

We write this as {{{4P[2]}}}.

The lower index 2 indicates the number of factors.  The upper index 4 indicates the first factor.
For example,  {{{8P[3]}}}  means "the number of permutations of 8 different things taken 3 at a time."  And {{{8P[3]}}}

For, there are 8 ways to choose the first, 7 ways to choose the second, and 6 ways to choose the third. 
In general,
<blockquote>{{{nP[k]}}} = {{{n*(n-1)*(n-2) * ellipsis * (n-k+1)}}} &nbsp;&nbsp;&nbsp;&nbsp;(k factors) </blockquote>

Factorial representation
<blockquote>{{{(8!)/(5!)=8*7*6}}}</blockquote>
{{{5!}}} is a factor of {{{8!}}}, and therefore the {{{5!}}} cancel.

Now, {{{8*7*6=8P[3]}}}. We see, then, that {{{8P[3]}}} can be expressed in terms of factorials as
<blockquote>{{{8P[3]=8!/(8-3)!=(8!)/(5!)}}}</blockquote>

In general, the number of arrangements -- permutations -- of n things take k at a time, can be represented as follows:
<blockquote>{{{nP[k]=n!/(n-k)!}}}</blockquote>

The upper factorial is the upper index of P, while the lower factorial is the difference of the indices.

EXAMPLE 3

Express  {{{10P[4]}}}  in terms of factorials.

Answer:10!/6!

The upper factorial is the upper index, and the lower factorial is the difference of the indices.  When the 6!'s cancel, the numerator
becomes {{{10*9*8*7}}}.
This is the number of permutations of 10 different things taken 4 at a time.

EXAMPLE 4

Calculate  {{{nP[n]}}}.

Answer:{{{n!/(n-n)!)}}}= {{{n!/0!}}} = {{{n!/1}}} = {{{n!}}}.

Hope this helps!