probability.
This topic has expert replies
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!=1/30kevinc628 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?
Success doesn't come overnight!
 Shalabh's Quants
 Master  Next Rank: 500 Posts
 Posts: 134
 Joined: 06 Apr 2012
 Thanked: 35 times
 Followed by:5 members
For I letter to go in wrong envelope; Probability = 29/30;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 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;
.
.
.
.
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.
Shalabh Jain,
eGMAT Instructor
eGMAT Instructor

 Junior  Next Rank: 30 Posts
 Posts: 29
 Joined: 25 Dec 2008
 Thanked: 2 times
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
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
with such wording of question we must assume:kevinc628 wrote:15 letters and 15 envelopes, What is the probability that no letter is placed in the right envelope?
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!
Success doesn't come overnight!

 Junior  Next Rank: 30 Posts
 Posts: 18
 Joined: 06 Apr 2012
 Thanked: 5 times
 Followed by:1 members
 GMAT Score:770
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.
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.
Kindly Use Thanks Button as liberally as you receive replies.
Mr. Smith
Mr. Smith
 GMATGuruNY
 GMAT Instructor
 Posts: 15532
 Joined: 25 May 2010
 Location: New York, NY
 Thanked: 13060 times
 Followed by:1897 members
 GMAT Score:790
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!) = 31 = 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!) = 124+1 = 9.
Total possible arrangements = 4! = 24.
P(no letter is placed in the correct envelope) = 9/24 = 3/8.
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!) = 31 = 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!) = 124+1 = 9.
Total possible arrangements = 4! = 24.
P(no letter is placed in the correct envelope) = 9/24 = 3/8.
Mitch Hunt
Private Tutor for the GMAT and GRE
[email protected]
If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.
Available for tutoring in NYC and longdistance.
For more information, please email me at [email protected].
Student Review #1
Student Review #2
Student Review #3
Private Tutor for the GMAT and GRE
[email protected]
If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.
Available for tutoring in NYC and longdistance.
For more information, please email me at [email protected].
Student Review #1
Student Review #2
Student Review #3