kaarthikvs wrote:Marsh has to put 6 different letters in 6 distinct envelopes. 1) If he puts exactly one of the letters in the correct envelope and the remaining 3 letters into wrong envelopes, in how many ways he can make such a mistake ? 2)In how many ways "No letter goes into its corresponding envelope"?

Check your source again.
If there are 6 letters and 6 envelopes, and exactly one of them goes to the correct envelope then (6 - 1) = 5 letters are left not 3.

kaarthikvs wrote:Marsh has to put 6 different letters in 6 distinct envelopes. 1) If he puts exactly one of the letters in the correct envelope and the remaining 5 letters into wrong envelopes, in how many ways he can make such a mistake ? 2)In how many ways "No letter goes into its corresponding envelope"?

It is unlikely that something of this kind you will see in GMAT.
This problem deals with the idea of Derangement, which is defined as "the number of arrangements of some different elements such that none of them are in their original place"

Number of derangement of n different elements are given by, D(n) = n![1 - 1/1! + 1/2! - 1/3! + 1/4! - ... (-1)^n/n!]

Question Number #1:
Let us assume only the 1st letter goes into its corresponding envelope and none of the rest 5 letters goes into their corresponding envelopes.

Number of ways such that none of the 5 letters goes into its corresponding envelope = D(5)
= 5![1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!]
= [5! - 5!/1! + 5!/2! - 5!/3! + 5!/4! - 5!/5!]
= [60 - 20 + 5 - 1]
= 44

Hence, for each of the 6 letters going into its corresponding envelope, the rest 5 letters can go into wrong envelopes in 44 ways.

Hence, number of ways such that exactly one of the letters go into in the correct envelope and the remaining five letters into wrong envelopes = 6*44 = 264


Question Number #2:
Number of ways such that no letter goes into its corresponding envelope = D(6)
= 6![1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!]
= [6! - 6!/1! + 6!/2! - 6!/3! + 6!/4! - 6!/5! + 6!/6!]
= [360 - 120 + 30 - 6 + 1]
= 265