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Tanya prepared 4 different letters to be sent to 4 different

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Intern
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Joined: 17 Sep 2017
Posts: 34

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

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New post 26 Nov 2017, 06:37
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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Joined: 03 May 2017
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Kudos [?]: 19 [1], given: 14

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

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New post 26 Nov 2017, 08:50
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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 [#permalink]

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New post 26 Nov 2017, 09:14
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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Tanya prepared 4 different letters to be sent to 4 different [#permalink]

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New post 26 Nov 2017, 09:50
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,

Kudos [?]: 19 [0], given: 14

Tanya prepared 4 different letters to be sent to 4 different   [#permalink] 26 Nov 2017, 09:50

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