i not getting the correct answer for this sum : In how many ways can five letters be placed in five envelopes such that exactly one letter goes into corresponding envelope?
Do you mean exactly one letter goes into each envelope?
You have 5 choices which letter to put into the first envelope, then 4 choices which to put in the second envelope, then 3, then 2, then 1. 5*4*3*2*1= 5!
This is exactly the same as the number of different ways to order 5 objects.
Plato's answer was assuming that each letter had a correct envelope and you were asking "how many ways can you put 5 letters in 5 evelopes if no letter goes into the correct envelope". I did not interpret your question that way.
Here is a very helpful website for derangements
Here is a formula: $\displaystyle D(n) = n!\sum\limits_{k = 0}^n {\frac{{( - 1)^k }}{{k!}}} $.
Here is a good approximation: $\displaystyle n \geqslant 4\, \Rightarrow \,D(n) \approx \frac{n!}{e}$.
$\displaystyle D(4)=\frac{4!}{e}\approx 8.8291$, so $\displaystyle D(4)=9$