Important Concepts on Permutation

Permutation

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-

ⁿP_r       =            n!/(n-r)!

Proof:     Say we have ‘n’ different things a₁, a₂……, a_n.

Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n-1

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

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:
ⁿP_r       = n (n-1)(n-2) --------------(n-r+1)

= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-1)] ----3.2.1

ⁿP_r = n!/(n-r)!

Number of permutations of ‘n’ different things taken all at a time is given by:-

ⁿP_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.

ⁿP_n =  n!

Concept. 

We have   ⁿP_r  =     n!/n-r

Putting r = n, we have :-

ⁿP_r  =   n! / (n-r)

But   ⁿP_n  =  n!

Clearly it is possible, only when  n!  = 1

Hence it is proof that     0! = 1

Note : Factorial of negative-number is not defined. The expression  –3! has no meaning.