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# One has four fair dice and rolls them together. What is the

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One has four fair dice and rolls them together. What is the [#permalink]

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30 Sep 2003, 16:46
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One has four fair dice and rolls them together. What is the probability of having all the numbers different? (for example, 1-2-3-4, 6-5-4-1, and so on)

One has six fair dice and rolls them together. What is the probability of having all the numbers different?
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01 Oct 2003, 07:02
1) 6*5*4*3/6^4 ( Number of favourable outcomes = 6C4*4! - i assume that dices are not identical, if they are identical dices than favourable outcomes = 6C4)
2) 6!/6^4
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02 Oct 2003, 11:12
Vicky wrote:
1) 6*5*4*3/6^4 ( Number of favourable outcomes = 6C4*4! - i assume that dices are not identical, if they are identical dices than favourable outcomes = 6C4)
2) 6!/6^4

i have some questions.

For 1, Does it matter that the Dice are identical....We are concerned with the outcomes...arent we? and why do you use combinations... 1,2 is different from 2,1.

For 2, why 6^4...was that a typo, we have six dice.
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02 Oct 2003, 19:57
For 2: yes it was a typo. thanks for correction.
For 1: For example: say if 3 dice are of colours: red, green & blue. Then getting r-1,g-2 & b-3 (1,2 & 3 are numbers on the dice) is different from getting r-2,g-3,b-2 or any other permutation with these numbers.
But if all are green dices then there is only one to get above numbers.
While i write this i also realize that outcomes will be different cause we have outcome on 1st, 2nd and 3rd dice (inherently different outcomes).
Praet thanks, i think we need to take permutations only..!!
what do u say..?
let us know what are correct answers.
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