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# As part of a game, four people each must secretly choose an

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VP
Joined: 25 Nov 2004
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As part of a game, four people each must secretly choose an [#permalink]

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21 Apr 2005, 23:45
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As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

Kudos [?]: 130 [0], given: 0

Manager
Joined: 05 Feb 2005
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Location: San Jose

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22 Apr 2005, 00:12
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

Total number of choice combos = 4x4x4x4
Number of choice combos where they choose diff numbers = 4!
(arrange 4 numbers into 4 empty slots = 4P4 = 4!)

Hence the answer = 4! / 256 = 3/32
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VP
Joined: 25 Nov 2004
Posts: 1481

Kudos [?]: 130 [0], given: 0

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25 Apr 2005, 08:46
no more.....................

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VP
Joined: 30 Sep 2004
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Location: Germany

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25 Apr 2005, 12:08
MA wrote:
no more.....................

number of different arrangements is 4! and total of 4^4 is possible. so it is 4!/4^4 = 3/32

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VP
Joined: 13 Jun 2004
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Location: London, UK
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25 Apr 2005, 19:36
Same approach as mckenna

total possibilities : 4^4
possibilities of 4 different numbers : 4*3*2*1 = 24

prob : 24/4^4 = 6/4^3 = 3/32

Kudos [?]: 51 [0], given: 0

25 Apr 2005, 19:36
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