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# PS: Combinatorics..

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Manager
Joined: 17 Dec 2008
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02 Mar 2009, 18:49
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Tanya prepares 4 letters to be sent to 4 addresses. For each letter she prepared an envelope with correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only one letter will be put into the envelope with correct address?

A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

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SVP
Joined: 07 Nov 2007
Posts: 1790

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Location: New York

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02 Mar 2009, 20:50
ConkergMat wrote:
Tanya prepares 4 letters to be sent to 4 addresses. For each letter she prepared an envelope with correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only one letter will be put into the envelope with correct address?

A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

You have 4 envelop and 4 letters--> Total no of ways = 4! =24

assume that 1234 -->only letter 1 will go to 1 other three will go into wrong enevelop

1-1 234 --> 432 234 -->342
only two possible solutions in this category
similarly when 2-2 other letters go to wrong address -->2 possible solutions.
similarly for 3-3 and 4-4
So No . of ways = 4C1 *2 = 8

p = 8/24 = 1/3
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Manager
Joined: 01 Mar 2009
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02 Mar 2009, 22:49
what is the C in 4C1*2=8?

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Manager
Joined: 02 Mar 2009
Posts: 134

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16 Mar 2009, 01:16
It stands for combination.

Suresh, I think there is a minor error.

When you were explaining the 1-1 bit, the two possibilities of going wrong are:

234 --> 342 NOT 234 --> 432

Anyway, this was a minor error.

Thanks

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Re: PS: Combinatorics..   [#permalink] 16 Mar 2009, 01:16
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