Where are the answer choices? Anyways, it could be as follow:

5 letters can go into 5 envelopes in 5! ways, i.e. 5! = 120, and there is just 1 combination in which each and every letter goes to its corresponding envelope, thus 120-1=119 for first condition

1 letter in its correct corresponding envelope and 4 others juggled, here we get 4! = 24 again we should subtract 1 combination in which all 4 letters can go correctly to their respective envelopes, thus 24-1 = 23 for Second condition.

To sum up, 119+23 = 142 ways.

Please, correct if I went awry

smanstar wrote:Q
In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?
2. Exactly 1 letter goes into its corresponding envelope?

Javoni wrote:Where are the answer choices? Anyways, it could be as follow:

5 letters can go into 5 envelopes in 5! ways, i.e. 5! = 120, and there is just 1 combination in which each and every letter goes to its corresponding envelope, thus 120-1=119 for first condition

1 letter in its correct corresponding envelope and 4 others juggled, here we get 4! = 24 again we should subtract 1 combination in which all 4 letters can go correctly to their respective envelopes, thus 24-1 = 23 for Second condition.

To sum up, 119+23 = 142 ways.

Please, correct if I went awry

smanstar wrote:Q
In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?
2. Exactly 1 letter goes into its corresponding envelope?

Well the 1st part requires that none of the letters go into the required envelope , what you considered is all the letters in going into its envelope, which is 1 way only , but there are many other combinations ( where 1 or 2 or 3 or 4 letter goes into its envelope) and subtract them from 120 ways.
hope you got it .

smanstar wrote:Q
In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Given 5 letters A, B, C, D and E:
Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44.
Total possible arrangements = 5! = 120.
P(no letter is in the correct position) = 44/120 = 11/30.

smanstar wrote:Q
In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?
2. Exactly 1 letter goes into its corresponding envelope?

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Let the correct ordering of the 5 letters be A-B-C-D-E.

P(A is correctly placed):
P(A is in the correct position) = 1/5. (Of the 5 positions, only 1 is correct.)

P(B, C, D and E are all incorrectly placed):
Total number of derangements = 4! (1/2! - 1/3! + 1/4!) = 12-4+1 = 9.
Total possible arrangements = 4! = 24.
P(no letter is in the correct position) = 9/24 = 3/8.

Since we want both events to happen, we multiply the probabilities:
1/5 * 3/8.
Since the correctly placed letter could be A, B, C, D or E -- yielding 5 options for the correctly placed letter -- the result above must be multiplied by 5:
5 * 1/5 * 3/8 = 3/8.

GMATGuruNY wrote:

smanstar wrote:Q
In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Given 5 letters A, B, C, D and E:
Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44.
Total possible arrangements = 5! = 120.
P(no letter is in the correct position) = 44/120 = 11/30.

hi thanks for the reply .
Is there any logical way to do it ???

And how to solve the Second part

smanstar wrote:

GMATGuruNY wrote:

smanstar wrote:Q
In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Given 5 letters A, B, C, D and E:
Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44.
Total possible arrangements = 5! = 120.
P(no letter is in the correct position) = 44/120 = 11/30.

hi thanks for the reply .
Is there any logical way to do it ???

And how to solve the Second part

The problems posted here are too complex for the GMAT.
For two similar problems that I solved without formulas, check here:

https://www.beatthegmat.com/probability-t55896.html
https://www.beatthegmat.com/probability-t121078.html