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There are 6 letters and 6 self addressed envelopes.What is

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There are 6 letters and 6 self addressed envelopes.What is [#permalink] New post 29 Jul 2003, 21:40
There are 6 letters and 6 self addressed envelopes.What is the probability that none is correctly placed.
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 [#permalink] New post 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
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 [#permalink] New post 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 1-83/720=637/720 :?: :?:

I think there's a mistake in my solution - want to understand what.

Also, want to understand your solution more.

Thanks!
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Re [#permalink] New post 30 Jul 2003, 05:06
The answer is 53/144.
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 [#permalink] New post 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 case--the 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!
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 [#permalink] New post 30 Jul 2003, 05:19
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.
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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


Last edited by AkamaiBrah on 30 Jul 2003, 05:47, edited 1 time in total.
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 [#permalink] New post 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
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 [#permalink] New post 30 Jul 2003, 05:59
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(n-1) and G(n-2). With a little bit of work, we can see that a general recursive formula for G(n) is:

G(n) = (n-1)*(G(n-1)+G(n-2)) 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).
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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


Last edited by AkamaiBrah on 30 Jul 2003, 06:12, edited 1 time in total.
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Re: Re [#permalink] New post 30 Jul 2003, 06:09
sanjaymishra wrote:
The answer is 53/144.


Sanjay

How did you get your answer?
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 [#permalink] New post 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(n-1) and G(n-2). With a little bit of work, we can see that a general recursive formula for G(n) is:

G(n) = (n-1)*(G(n-1)+G(n-2)) 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!!!
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Re [#permalink] New post 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/n-C(n,2)1/n*1/n-1 .......(-1)^n-1*C(n,n)*1/n!
=1-1/2!+1/3!..... +(-1)^n-1*1/n!

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Re: Re [#permalink] New post 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/n-C(n,2)1/n*1/n-1 .......(-1)^n-1*C(n,n)*1/n!
=1-1/2!+1/3!..... +(-1)^n-1*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.
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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

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Re: There are 6 letters and 6 self addressed envelopes.What is [#permalink] New post 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 * (6-1)! 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 (6-2)!
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.
Re: There are 6 letters and 6 self addressed envelopes.What is   [#permalink] 29 Oct 2012, 03:38
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