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Tanya prepared 4 different letters to be sent to 4 different addresses

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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 26 Nov 2017, 07:37
Can anyone please tell me why the solution has nothing to do with the given " 4 letters have to be send to 4 dif addresses" information?
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 26 Nov 2017, 09:50
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lichting wrote:
Can anyone please tell me why the solution has nothing to do with the given " 4 letters have to be send to 4 dif addresses" information?


Happy to help

Here is the question:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

The solution has all to do with 4 letters sent to 4 addresses, but we are specifically concerned about mismatching the letters and their corresponding addresses, albeit with one correct letter/address pair. However, we can choose any pair to be right and any 3 pairs to be wrong. I think your confusion might come from not seeing the letter/address as a pair but rather as separate entities.

All you had to do is find the number of either choosing a right pair out of four or 3 wrong pairs out of four. These are 4C1 and 4C3 respectively which are both 4. Now we have 2*4 ways with our constraint out of a total of 4! ways of choosing a pair of letter and address.

Hence the probability of only one correct pair of letter/address is
\(\frac{8}{24} = \frac{1}{3}\)

Now if you want to go the probability route, you can say:

The component probabilities of a correct pair with 3 wrong pairs are

\(\frac{1}{4}\) (1st pair correct)
\(\frac{2}{3}\)( 2nd pair wrong, one could be right, 2 definitely wrong)
\(\frac{1}{2}\)( 3rd pair wrong, one could be right,1 definitely wrong)
1 (4th pair wrong is absolute)

Note that its good to keep in mind that any pair could be wrong or right, hence we have 4 ways (4C1) of our iteration i.e any of the pairs could occupy the first correct slot. Also, we can start this iteration by starting with 1st wrong.....4th right, regardless we had arrive at the same answer

From above our probability is\(\frac{1}{4} *\frac{2}{3}*\frac{1}{2}*1 = \frac{1}{12}\)
Now we multiply by the number of ways i.e \(\frac{1}{12}*4 = \frac{1}{3}\)
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 26 Nov 2017, 10:14
Quote:
I think your confusion might come from not seeing the letter/address as a pair but rather as separate entities.

Yes!! This's exactly what I used to think. But after you boiled it down, I figured it out. and the latter approach seems much more easier for me to understand than the shorter way.
Thank you!
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 26 Nov 2017, 10:50
lichting wrote:
Quote:
I think your confusion might come from not seeing the letter/address as a pair but rather as separate entities.

Yes!! This's exactly what I used to think. But after you boiled it down, I figured it out. and the latter approach seems much more easier for me to understand than the shorter way.
Thank you!


Happy to help. There is also a "handshake" combinatorics question on one of the GMAT prep exams that exploited the pairing concept. You might want to check it out.

Best,
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 04 May 2018, 10:03
I think that since there cannot be only one envelope with the wrong letter (if one envelope has a wrong letter, then another one must also have a wrong letter) the number of possible combinations reduces to 12:

A B C D
g g g g
g g x x
g x g x
g x x g
x g g x
x g x g
x x g g
x x x x
g x x x - Only one envelope has the right letter
x g x x - Only one envelope has the right letter
x x g x - Only one envelope has the right letter
x x x g - Only one envelope has the right letter

So out of the 12 possible combinations, only 4 have the right letter in the right envelope. This yields 4/12 = 1/3. :dazed
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 14 Dec 2018, 01:56
Similar pattern question to practice
https://gmatclub.com/forum/each-of-four ... 01553.html
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Please mention my name in your valuable replies.

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Re: Tanya prepared 4 different letters to be sent to 4 different addresses  [#permalink]

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New post 07 Jan 2019, 07:29
Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability:

A. That no letter will be put into the envelope with its correct address?

B. That all letters will be put into the envelope with its correct address?

C. That only 1 letter will be put into the envelope with its correct address?

D. That only 2 letters will be put into the envelope with its correct address?

E. That only 3 letters will be put into the envelope with its correct address?

F. That more than one letter will be put into the envelope with its correct address?

G. That more than two letters will be put into the envelope with its correct address?

Posting Solution for all of the above questions.

A) No letter in the correct envelope: If you count manually, there are nine ways of getting all letters in the wrong envelope.
No of ways to arrange 4 different letters in 4 different envelopes = 4! ways = 24
Probability = 9/24

B) All letters in the correct envelope: There is only 1 way of arranging all letters in the correct envelope.
Hence, Probability = 1/24

C) Only 1 letter in the correct envelope: No of ways to select 1 from 4 letters to put in the correct envelope = 4C1 = 4 ways.
Lets say Letter 1 is in the correct envelope. Letters 2-3-4 can be arrange in the wrong envelope in only 2 ways i.e. 3-4-2 or 4-2-3.
Possibilities = 4*2 = 8 ways
Prob = 8/24 = 1/3

D) Only 2 letters in the correct envelope: No of ways to select 2 letters from 4 letters = 4C2 = 6 ways.
Lets say letter 1 & letter 2 are in the correct envelope. There is only 1 way of putting letters 3 and 4 in the wrong envelopes.
Possibilities = 6*1 = 6
Prob = 6/24 =1/4

E) Only 3 letters in the correct envelope: Not Possible. If 3 letters are in the correct envelope, the 4th must also be in the correct envelope.

F) More than 1 letter in correct address:
[Note: No of ways all 4 letters in correct envelope = 1. Hence, Prob of all 4 letters in correct envelope = 1/24]
Prob of 2 letters + Prob of 3 letters + Prob of all 4 letters = 1/4 + 0 + 1/24 = 7/24

G) More than 2 letters in correct envelope:
Prob of 3 letters + Prob of all 4 letters = 0 + 1/24 = 1/24
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Re: Tanya prepared 4 different letters to be sent to 4 different addresses   [#permalink] 07 Jan 2019, 07:29

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