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# I know this one has been discussed many times, but I would

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Manager
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I know this one has been discussed many times, but I would [#permalink]

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28 Apr 2007, 08:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I know this one has been discussed many times, but I would appreciateif you remind me how to solve this. Thanks!
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28 Apr 2007, 09:32
There are 4 letters and 4 envelopes.

Probability of putting 1st letter in correct envelope (And others in wrong envelopes):
P(1st) = 1/4 * 2/3 * 1/2 = 1/12

So, probability of putting only one letter in correct envelope = 4 x P(1st) = 1/3
Manager
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28 Apr 2007, 11:07
Sorry, but I did not get the explanation above. Can you be more specific?
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28 Apr 2007, 11:38
P(1st) = Only E1 has correct letter (L1). Other envelopes have wrong letters
= P(picking L1 for E1) X P(then not picking L2 for E2) X P(then mismatches in other two envelopes)
= 1/4 * 2/3 * 1/2 = 1/12

Since P(1st) = P(2nd) = P(3rd) = P(4th)
Director
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28 Apr 2007, 12:23
did it in a different way ..

Total number of ways 4 letters can be placed in 4 env = 4! = 24

Now, to find the number of ways with the constraint stated :

# Choices for the correct letter to choose from 4 env = 1

# choice for each of the other letters = 2 (1 env is already taken by correct letter and it can choose only 2 of the remaining 3 )

so total number of ways = 1 x 2 x 2 x 2 = 8

hence probability = 8/24 = 1/3
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28 Apr 2007, 14:31
Sorry, but I did not get the explanation above. Can you be more specific?

Total 4! = 24

one letter to one envelope:

1 2 3 4
1 3 4 2
1 4 2 3

If 1 fits - you have two different ways to move the rest (and so on if 2,3,4 fits so 2x4 = 8)

so 8/24 = 1/3

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