February 24, 2019 February 24, 2019 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. February 23, 2019 February 23, 2019 07:00 AM PST 09:00 AM PST Learn reading strategies that can help even nonvoracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 31 May 2003
Posts: 6
Location: New Jersey

There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
29 Jul 2003, 21:40
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 1 sessions
HideShow timer Statistics
There are 6 letters and 6 self addressed envelopes. What is the probability that none is correctly placed.



SVP
Joined: 03 Feb 2003
Posts: 1556

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 01:19
We have 6 envelopes (E) and 6 letters (L)
The matrix is
E E E E E E
L L L L L L
Total outcomes = 6 items in 6 places= 6!
Favorable: 6C1 (each of 6 letters) * 5C1 (takes one of 5 wrong envelopes) = 6*5=30
P=30/720=1/24



Manager
Joined: 24 Jun 2003
Posts: 141
Location: India

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 02:47
stolyar wrote: We have 6 envelopes (E) and 6 letters (L) The matrix is E E E E E E L L L L L L
Total outcomes = 6 items in 6 places= 6! Favorable: 6C1 (each of 6 letters) * 5C1 (takes one of 5 wrong envelopes) = 6*5=30
P=30/720=1/24
Stolyar,
I'm not sure if I'm correct. But here's another solution.
P(none is placed in right envelope) = 1  P(atleast one is placed in the right envelope)
P(atleast one is placed in the right envelope) = (6C1+6C2+6C3+6C4+6C5+6C6) / 6! = 83/720
Therefore the reqd prob is 183/720=637/720
I think there's a mistake in my solution  want to understand what.
Also, want to understand your solution more.
Thanks!



Intern
Joined: 31 May 2003
Posts: 6
Location: New Jersey

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 05:06
The answer is 53/144.



SVP
Joined: 03 Feb 2003
Posts: 1556

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 05:06
I myself do not know whether I am right. Your answer seems to be too large, close to 1. But we deal with an exceptional casethe probability should be small.
I base my solution on the assumption that each letter has 5 chances to take a wrong envelope and the only chance to take a right one.
BTW, the probability that all the letters will be placed correctly is 1/6!



GMAT Instructor
Joined: 07 Jul 2003
Posts: 740
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
Updated on: 30 Jul 2003, 05:47
sanjaymishra wrote: There are 6 letters and 6 self addressed envelopes.What is the probability that none is correctly placed.
I believe the answer is 265/720 = 53/144.
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
Originally posted by AkamaiBrah on 30 Jul 2003, 05:19.
Last edited by AkamaiBrah on 30 Jul 2003, 05:47, edited 1 time in total.



Manager
Joined: 24 Jun 2003
Posts: 141
Location: India

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 05:42
AkamaiBrah wrote: sanjaymishra wrote: There are 6 letters and 6 self addressed envelopes.What is the probability that none is correctly placed. Consider this: I believe the answer is 265/720 = 53/144.
Akamai,
Please elaborate on your answer



GMAT Instructor
Joined: 07 Jul 2003
Posts: 740
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
Updated on: 30 Jul 2003, 06:12
prashant wrote: AkamaiBrah wrote: sanjaymishra wrote: There are 6 letters and 6 self addressed envelopes.What is the probability that none is correctly placed. Consider this: I believe the answer is 265/720 = 53/144. Akamai, Please elaborate on your answer
Solution:
First of all, IMO, although this problem looks basic, in reality this problem is WAY WAY WAY too difficult for the GMAT and will stump even graduate math students (it took me about 30 minutes).
We are looking for the solution of G(6) where G(n) = number of ways n items can be permuted such that they none of them match their original position. This is the same as having a deck of n cards numbered from 1 to n, then counting the ways you can shuffle the deck without any cards matching its position.
Consider this: it is obvious that when n = 1, G(n) must be ZERO, and when n = 2, G(n) = 1 (i.e., {2,1}), and when n = 3, G(n) = 2 (i.e., {2,3,1},{3,1,2}). By drawing out a probability tree, you will discover that G(4) = 9, and with a LOT of work and patience, G(5) = 44.
If you carefully watch how the tree is growing, you may notice that the results of G(n) refer back to the results of G(n1) and G(n2). With a little bit of work, we can see that a general recursive formula for G(n) is:
G(n) = (n1)*(G(n1)+G(n2)) where n > 1 and G(1) = 0 and G(2) = 1.
If you build a table, then G(6) comes out to 5*(44 + 9) = 265. Since there are 6! or 720 ways to stuff the envelopes, the probabiltiy of getting them all wrong is 265/720 = 53/144.
(if you don't believe me, run a simulation).
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
Originally posted by AkamaiBrah on 30 Jul 2003, 05:59.
Last edited by AkamaiBrah on 30 Jul 2003, 06:12, edited 1 time in total.



Manager
Joined: 24 Jun 2003
Posts: 141
Location: India

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 06:09
sanjaymishra wrote: The answer is 53/144.
Sanjay
How did you get your answer?



Manager
Joined: 08 Apr 2003
Posts: 140

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 15:01
AkamaiBrah wrote: prashant wrote: AkamaiBrah wrote: sanjaymishra wrote: There are 6 letters and 6 self addressed envelopes.What is the probability that none is correctly placed. Consider this: I believe the answer is 265/720 = 53/144. Akamai, Please elaborate on your answer Solution: First of all, IMO, although this problem looks basic, in reality this problem is WAY WAY WAY too difficult for the GMAT and will stump even graduate math students (it took me about 30 minutes). We are looking for the solution of G(6) where G(n) = number of ways n items can be permuted such that they none of them match their original position. This is the same as having a deck of n cards numbered from 1 to n, then counting the ways you can shuffle the deck without any cards matching its position. Consider this: it is obvious that when n = 1, G(n) must be ZERO, and when n = 2, G(n) = 1 (i.e., {2,1}), and when n = 3, G(n) = 2 (i.e., {2,3,1},{3,1,2}). By drawing out a probability tree, you will discover that G(4) = 9, and with a LOT of work and patience, G(5) = 44. If you carefully watch how the tree is growing, you may notice that the results of G(n) refer back to the results of G(n1) and G(n2). With a little bit of work, we can see that a general recursive formula for G(n) is: G(n) = (n1)*(G(n1)+G(n2)) where n > 1 and G(1) = 0 and G(2) = 1. If you build a table, then G(6) comes out to 5*(44 + 9) = 265. Since there are 6! or 720 ways to stuff the envelopes, the probabiltiy of getting them all wrong is 265/720 = 53/144. (if you don't believe me, run a simulation).
One of the most interesting problem i have come across!!!
Bravo!!!! Akamai!!!



Intern
Joined: 31 May 2003
Posts: 6
Location: New Jersey

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 18:56
AkamaiBrah
Can you please explain it more elaborately.
I saw a similar question: An inefficient secretary places n different letters into n differently addressed envelopes at random.Find the probability that atleast one of the letters will arrive at the proper destination.
P=C(n,1)1/nC(n,2)1/n*1/n1 .......(1)^n1*C(n,n)*1/n!
=11/2!+1/3!..... +(1)^n1*1/n!
Sanjay



GMAT Instructor
Joined: 07 Jul 2003
Posts: 740
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
30 Jul 2003, 19:24
sanjaymishra wrote: AkamaiBrah
Can you please explain it more elaborately.
I saw a similar question: An inefficient secretary places n different letters into n differently addressed envelopes at random.Find the probability that atleast one of the letters will arrive at the proper destination. P=C(n,1)1/nC(n,2)1/n*1/n1 .......(1)^n1*C(n,n)*1/n! =11/2!+1/3!..... +(1)^n1*1/n!
Sanjay
I don't have time to derive the result you got, but if you plug in the numbers, it is consistent with my solution for any n. (The solution for this problem is 1 minus the probability for the original problem).
While this is an interesting problem, it is not a GMAT problem, nor does it yield any insights to help with GMAT problems, so I will leave it at this.
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



Intern
Joined: 29 Oct 2012
Posts: 1

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
29 Oct 2012, 03:38
HI Prashant, You are getting 637/720 since you are counting repeated combinations. Here goes the solution 6 cards need to be placed in 6 envelops such that not even one card is in the right envelop. Let us assume that atleast one card is in the right envelop. Number of ways in which atleast one card is placed in the right envelop is 6C1 * (61)! ways. [We select one card and place it in its right envelop and arrange the rest of the 5 cards in the remaining 5 envelops.] This should be subtracted from 6! However in the above cases where atleast one card is placed in the right envelop, the cases where exactly two cards are placed in the right envelop are counted twice. Since we cannot find the number of cases where exactly two cards are placed in the correct envelops, we add back the cases where atleast 2 cards are placed in the right envelops which is 6C2 (62)! While adding the above cases, all the cases where exactly 3 cards are placed in the right envelops are also added back which should be subtracted. Thus we see a pattern of subtractions and additions . Thus the number of ways is 6!6C1(5!)+6C2(4!)6C3(3!)+6C4(2!)+6C5(1!)+6C6(0!)= 256 ways. Thus the probability that none of them are in the correct envelops is 256/720 ways.



NonHuman User
Joined: 09 Sep 2013
Posts: 9895

Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
Show Tags
29 Jan 2019, 23:04
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: There are 6 letters and 6 self addressed envelopes. What is the probab
[#permalink]
29 Jan 2019, 23:04






