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$

Fundamental Principle of counting

Addition rule

: 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 $0! = 1$

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 =$\frac {n!}{p_1! p_2!....p_r!}$

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