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

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CEO
Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 704 [0], given: 781

One has four fair dice and rolls them together. What is the [#permalink]  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?
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
Followers: 1

Kudos [?]: 4 [0], given: 0

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
CEO
Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 704 [0], given: 781

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.
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
Followers: 1

Kudos [?]: 4 [0], given: 0

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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