let 5 envelopes be E1,E2,E3,E4,E5. now first letter can go to any of 5 envelopes in 5 ways. second letter can go to any of the 4 remaining envelopes in 4 ways. third letter can go to remaining 3 envelopes in 3 ways. fourth letter can go to remaining 2 envelopes in 2 ways. and finally last envelopes has only 1 ways.
so when no restrictions are applied. these 5 letters can arrange themselves in 5 envelopes in 5.4.3.2.1=120 ways

(we can also think of this as arrangements of 5 students in 5 vacant spaces.)

now, obviously there is only one correct envelope designated for each letter. thus the number of ways in which all letters will be correctly posted=1

thus required probability = 1/120

A-B-C-D-E (envelopes)
a-b-c-d-e (letters)

5 letters (a, b, c, d, and e) can be assigned to 5 envelopes (arranged) in 5! ways, out of which only one will be correct, thus the probability is 1/5! = 1/120.

Answer: C.
_________________

Correct option : C - (1/120)
Method: Permutation (arragement of 5 letters in selection 5 envolope)
Formula : nPr
Given : Letter : 5, Envolope : 5
Probability P = (1 / nPr) = 1 / 5P5 = [1/ (5x4x3x2x1)] = 1 / 120 - Correct Option C