**Permutation**

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

^{n}P

_{r}= n!/(n-r)!

**Proof**: Say we have ‘n’ different things a

_{1}, a

_{2}……, 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:

^{n}P_{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

^{n}P

_{r}= n!/(n-r)!

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.

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 (n-1) (n-2) ------ 2.1.

^{n}P

_{n}= n!

Concept.

We have

^{n}P_{r}= n!/n-rPutting r = n, we have :-

^{n}P

_{r}= n! / (n-r)

But

^{n}P_{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.

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