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# one more probablity from GMATPrep

Author Message
Manager
Joined: 16 Jan 2008
Posts: 105
Followers: 1

Kudos [?]: 11 [0], given: 6

one more probablity from GMATPrep [#permalink]

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11 Oct 2008, 07:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Guys,

can someone get me the breakup of this?

Thanks again..
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Manager
Joined: 10 Aug 2008
Posts: 76
Followers: 1

Kudos [?]: 13 [1] , given: 0

Re: one more probablity from GMATPrep [#permalink]

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11 Oct 2008, 11:59
1
KUDOS
lets proceed in this way.
Assume 4 envelops is E1,E2,E3 & E4
& 4 letters are L1,L2,L3 & L4

Total Number of way 4 letters can put in 4 Envelops
4! = 24

Now let suppose L1 put in correct envelope E1 than only 2 ways the remaing 3 letters can put in 3 envelops so NOT any letter put in correct envelope.

E1 E2 E3 E4
-------------------
L1 L4 L2 L3
-> In E2 only 2 letters can be put either L4 or L3, if L4 than only L2 can be put in E3
L1 L3 L4 L2
-> If L3 put into E2 than only L4 can be put in E3, because No Other envelope can get a correct letter.

So only 2 ways is [possible if E1 gets L1. This is true for each envelope-letter combination.
So number of ways in which only 1 letter put into correct envelope is
2x4 = 8

Probability 8/24 = 1/3
Re: one more probablity from GMATPrep   [#permalink] 11 Oct 2008, 11:59
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