 # Permutation

## What is factorial?

The product of first ‘n’ natural numbers is denoted by $n!$.
$n! = n(n-1) (n-2) ......3.2.1$
Example
$5! = 5 \times 4 \times 3 \times 2 \times 1 =120$
Note $0! =1$
Proof
$n! =n \times (n-1)!$
or $(n-1)! = \frac {n!} {n}$
Putting n = 1, we have
$0! = \frac {1!}{1}$
or $0! = 1$
Example
$(n+2)!=2550 \times n!$
Solution
$(n+2)!=2550 \times n!$
$(n+2)(n+1)n!=2550 \times n!$
Canceling n!
$(n+2)(n+1)=2550$
$n=49$

## Fundamental Principle of counting

: If an job can be performed in ‘n’ ways, & another job can be performed in ‘m’ ways then either of the two jobs can be performed in (m+n) ways. This rule can be extended to any finite number of jobs.

Example
In a class 12 , 20 students are from Maths streams  while 30 students are from Biology stream.
Teacher has to choose 1 student either from Maths stream or Biology stream to represent the class
What ways the selection can be made?
Solution:
20 students are from Maths:
Students can be chosen from Maths  stream in 20 ways
30 students are from Biology stream
Students can be chosen from Biology  stream in 30 ways
So by the Fundamental Principle of  addition rule, the selection of 1 student from either of two streams can be done (20+30)=50 ways

### Multiplication Rule

: If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in $m \times n$ ways. It can be extended to any finite number of operations
Example:
In a class 12 , 20 students are from Maths streams  while 30 students are from Biology stream.
Teacher has to choose 2 student one from  Maths stream  and one from  Biology stream to represent the class in Quiz competition
What ways the selection can be made?
Solution:
20 students are from Maths:
1 Students can be chosen from Maths  stream in 20 ways
30 students are from Biology stream
1Students can be chosen from Biology  stream in 30 ways
So by the Fundamental Principle of  Multiplication rule, the selection of 1 student from Maths  streams  and 1 students from Biology stream  can be done $(20 \times 30)=600$ ways

## Permutation

Permutation means arrangement of things. The word arrangement is used, if the order of things is considered

### Theorem 1

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-
$^{n}P_r=\frac { n!}{ (n-r)!}$
Proof:
Say we have ‘n’ different things $a_1 , a_2, ... , a_n$
1. Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = $n-1$
2. So the second-place can be filled-up in $(n-1)$ ways. Now number of things left after filling-up the first and second places = $n - 2$
3. Now the third place can be filled-up in $(n-2)$ ways. Thus number of ways of filling-up first-place = $n$
Number of ways of filling-up second-place = $n-1$
Number of ways of filling-up third-place = $n-2$
Number of ways of filling-up r-th place = $n - (r-1) = n-r+1$
By multiplication – rule of counting, total no. of ways of filling up, first, second --  rth-place together :-
$n (n-1) (n-2)..... (n-r+1)$
Hence:
$^{n}P_r = n (n-1)(n-2).....(n-r+1)$
$= \frac {[n(n-1)(n-2)...(n-r+1)] [(n-r)(n-r-1)...3.2.1.]}{ [(n-r)(n-r-1)...3.2.1]}$
$^{n}P_r=\frac { n!}{ (n-r)!}$ ### Theorem 2:

Number of permutations of ‘n’ different things taken all at a time is given by:-
$^{n}P_n= n!$
Proof
Now we have ‘n’ objects, and n-places.
Number of ways of filling-up first-place  = $n$
Number of ways of filling-up second-place = $n-1$
Number of ways of filling-up third-place  = $n-2$
Number of ways of filling-up r-th place, i.e. last place =1
Number of ways of filling-up first, second, --- n th place
$= n (n-1) (n-2) .... 2.1.$
$^{n}P_n= n!$

To prove
$0! = 1$
We have  $^{n}P_r=\frac { n!}{ (n-r)!}$
Putting r = n, we have :-
$^{n}P_n =\frac { n!}{ (n-n)!}= \frac {n!}{0!}$
But $^{n}P_n= n!$
Clearly it is possible, only when $0! = 1$
Hence it is proof that $0! = 1$

Example
How many different signals can be made by 4 flags from 6-flags of different colours?
Solution :
Number of ways taking 4 flags out of 6-flags = $^{6}P_4=\frac { 6!}{ (6-4)!}$
$=6 \times 5 \times 4 \times 3 = 360$

Example
How many words can be made by using the letters of the word “ZEBRA” taken all at a time?
Solution
There are ‘5’ different letters of the word “ZEBRA”
Number of Permutations taking all the letters at a time  = $^{n}P_n$
$=5! = 5 \times 4 \times 3 \times 2 \times 1=120$

## Restricted – Permutations

### Theorem -1

Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement
$r ^{n}P_{r-1}$
Proof:
When one particular thing is always taken, then that can arranged in r ways
Now we have to arrange the remaining $r-1$ places with $n-1$ objects,so permutations will be $^{n-1}P_{r-1}$ . Now by principle of counting ,both of these can be arrange in $r ^{n}P_{r-1}$

### Theorem - 2

Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed:
$= ^{n-1}P_{r-1}$
Proof:
When one particular thing is fixed ,then that is the only arrangement for that.Now we have to arrange the remaining $r-1$ places with $n-1$ objects,so permutations will be $^{n-1}P_{r-1}$ . Now by principle of counting ,both of these can be arrange in $^{n-1}P_{r-1}$

### Theorem -3

Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken
$= ^{n-1}P_{r}$
Proof:
When one particular thing is never taken then  we have to arrange the r- places with n-1 objects,so permutations will be $= ^{n-1}P_{r}$

## Another Theorem of Permutation

### Theorem

There  are n subjects of which $p_1$ are alike of one kind; $p_2$ are alike of another kind; $p_3$ are alike of third kind and so on and pr are alike of rth kind,
such that $(p_1 + p_2 + ... p_r) = n$.
Then, number of permutations of these n objects is =$\frac {n!}{p_1! p_2!....p_r!}$

## Solved Examples

Question-1
How many words can be formed with the letters of the word ‘TIGER’ when:
1. T’ and ‘R’ occupying end places.
2. G’ being always in the middle
3. Vowels occupying odd-places
4. Vowels being never together.
5. Find the number of words formed with out any restriction
Solution
1. When ‘T’ and ‘R’ occupying end-places
IGE. (TR)
Here (TR) are fixed, hence I,E,G can be arranged in  3! ways
But (T,R) can be arranged themselves is 2! ways.
Total number of words =$3! \times 2! = 12$ ways.
2. When ‘G’ is fixed in the middle
T.I.(G), E.R
Hence four-letter T.I.E.R. can be arranged in  4!   i.e 24 ways.
3. Two vowels (I,E) can be arranged in the odd-places (1st, 3rd and 5th) =  3! ways.
And three consonants (T,G,R) can be arranged in the remaining places               =   3 !   ways
Total number of ways= $3! \times 3! = 36$ ways.
4. Total number of words   =   5!   =    120
If all the vowels come together, then we have: (I.E.), T,R,G
These can be arranged in    4!    ways.
But (I,E.) can be arranged themselves in   2! ways.
Number of ways, when vowels come-together  = $4! \times 2!= 48$ ways
Number of ways, when vowels being never-together
$= 120-48 =72$ ways.
5. Total number of words   =   5!   =    120
Question-2
Find the numbers of words formed  by permuting all the letters of the following words
1. INDIA
2. RUSSIA
3. AUGUST
4. ARRANGE
Solution
1. In INDIA,  There are 2 I’s and all other are distinct
So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$
$=\frac {5!}{2!} =60$ words
2. In RUSSIA,  There are 2 S’s and all other are distinct
So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$
$=\frac {6!}{2!} =360$ words
3. In AUGUST,  There are 2 U’s and all other are distinct
So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$
$=\frac {6!}{2!} =360$ words
4)         In ARRANGE,  There are 2 R’s  and 2A’s and all other are distinct
So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$
$=\frac {7!}{2! \times 2!}= =1260$ words