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

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As part of a game, four people each must secretly choose an [#permalink]  02 Dec 2004, 12:42
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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?
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4!/4^4 = 3/32
prob of having 4 different numbers drawn by 4 different people where order counts: 4!
total # of outcomes for each person to draw 1 of 4 different numbers: 4^4
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Sole combination is (1,2,3,4) so 4! ways to allocate these 4 numbers.
Each person has 4 choices so total number of picks = 4^4

4!/4^4 = 6.4/(4.64)=6/64=3/32 id est 1/11 = 0,090909...
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OA [#permalink]  02 Dec 2004, 15:17
OA is 9%..thanks for the explanations
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OA   [#permalink] 02 Dec 2004, 15:17
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