kevinc628 wrote:We have 30 letters addressed to 30 different people, and 30 envelopes, one addressed to each person. Each letter is placed in an envelope at random. What is the probability that no letter is placed in the right envelope?

There are many possible scenarios: all letters in right envelopes, one is in right and remaining are in wrong envelopes, etc. The total outcomes will be 30! and favorable outcomes will be 29! as the first envelope may be filled with all choices except for one right letter, the second with 28 choices and so on and on ... Hence the required probability will be 29!/30!=1/30

kevinc628 wrote:30 letters and 30 envelopes, Each letter is placed in an envelope at random. What is the probability that no letter is placed in the right envelope?

For I letter to go in wrong envelope; Probability = 29/30;

For II letter to go in wrong envelope; Probability = 28/29;

For III letter to go in wrong envelope; Probability = 27/28;

For IV letter to go in wrong envelope; Probability = 26/27;
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For 29th letter to go in wrong envelope; Probability = 1/2;

For 30th letter to go in wrong envelope; Probability = 1;

Total probability = 29/30.28/29.27/28..........2/3.1/2.1 = 1/30.

Because the letter cannot be placed the second time in the same slot, so there are 14 possibilities of letters not put into the right envelopes. Probability is 14/15
The first letter is secured, so there are 14 possibilities left. Second letter cannot be placed the second time in the same slot, so there 13 possibilities of letter not out into the right envelopes. Probability is 13/14
Keep filling up the probability for the 15th envelope, probability is 1/2
So we have
14/15*13/14*12/13.....*2/3*1/2

Cancel up the numerator and denominator because it is in pair so we have

1/15

kevinc628 wrote:15 letters and 15 envelopes, What is the probability that no letter is placed in the right envelope?

with such wording of question we must assume:
1) each letter is placed in one envelope
2) the letters are placed at random
3) the letters placed in different envelopes are unique

if you have edited the question in haste of knowing different possibilities of 15 letter combination sets, then your job yet has to be continued, i mean editing question. With such wording the answer to your question is 0 - no action required!

Ok heres what I understood from your question:
You have 15 letters and 15 corresponding envelops. The 15 letters are placed in the 15 envelops, one in each. You want to find the probability that that each letter is placed in a wrong envelop.
If this is the question, it is way beyond the scope of GMAT. at least for 15 envelops.
The flaw in the method suggested by pemdas is that after he assumes that after having 29 choices for the first assignment, he'll have 28 choices for the second and so on. But actually the number of choices for each assignment depends on previous assignments. So if the first assignment is a letter for the second envelop then you'll have 29 choices and 28 otherwise. This complexity increases as you start counting the choices for successive assignments.

If you are really interested, I found this link online where theres a very detailed discussion about such problems:
https://www.physics.harvard.edu/academic ... /sol16.pdf


Coming back to GMAT, They can maybe ask for a more manageable number like say 3 letters and 3 envelops.
Then you have total 3! ways =6 ways of allotting the letters.
The count the number of ways of having a wrong letter in each envelop
here are the 2 ways,
{L2E1, L3E2, L1E3}, {L3E1, L1E2, L2E3}
hence probability = 2/6 = 1/3

similarly for 4 envelops:
Total ways: 4! = 24 ways
all wrong assignments 9 ways:
{L2E1, L1E2, L4E3, L4E4}, {L2E1, L3E2, L4E3, L1E4}, {L2E1, L4E2, L1E3, L1E4},
{L3E1, L1E2, L4E3, L2E4}, {L3E1, L4E2, L1E3, L2E4}, {L3E1, L4E2, L2E3, L1E4},
{L4E1, L1E2, L2E3, L3E4}, {L4E1, L3E2, L1E3, L2E4}, {L4E1, L3E2, L2E3, L1E4}
Hence probability: 9/24 = 3/8
Another approach of this problem is to count the number of configurations with atleast one correct assignment which 4 in the case of 3 envelops and 15 in the case of 4 envelops. But this I feel is more tedious

As you can see counting becomes significantly tedious as the number of letters increases.

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!)

Question: What is the probability that no letter is placed in the correct envelope?

3 letters and 3 envelopes:
Number of derangements = 3! (1/2! - 1/3!) = 3-1 = 2.
Total possible arrangements = 3! = 6.
P(no letter is placed in the correct envelope) = 2/6 = 1/3.

4 letters and 4 envelopes:
Number of derangements = 4! (1/2! - 1/3! + 1/4!) = 12-4+1 = 9.
Total possible arrangements = 4! = 24.
P(no letter is placed in the correct envelope) = 9/24 = 3/8.