Roll 12 different dice. How many will have at least each 1,2,3,4,5,6 occurring?

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The discussion focuses on calculating the probability of rolling 12 different dice and obtaining at least one of each number from 1 to 6. Each die has 6 sides, leading to a total of 6^12 possible outcomes when rolling 12 dice. The challenge is to determine how many of these outcomes include at least one occurrence of each number. This is a combinatorial probability problem that requires understanding of permutations and the principle of inclusion-exclusion to arrive at the correct solution.

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Homework Statement


12 different dice are rolled. How many outcomes will have at least one of each number 1,2,3,4,5,6 occurring?

The Attempt at a Solution


I don't even know where to go because I really don't know what the question is asking. Does it mean how many rolls (where each roll is rolling 12 dice) must be made to get at least one of each number?

Thanks
 
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You roll 12 dice.

Each (standard) die has 6 sides. You roll one die, it has 6 different outcomes. You then roll the second die, it has 6 outcomes. For the first two die, you could have 1-1, 1-2, 2-1, 2-2, 3-1, 3-2, 3-3, 2-3, 1-3, etc.

So, after rolling all 12 dice, you have 12*6 possibilities in total.

In how many of those possibilites will you have at least one die being a 1, one being a 2, one being a 3, 4, 5, 6..?
 
It seems to me like a probability question...

Since there are six numbers possible on a dice, and you have have 12 dices, the chances for any of those numbers(1,2,3,4,5, or 6) to come on a single dice is 1/6(<-- fraction). Can you figure out the rest?
 
pretty close goldenwing, 2 minutes difference(between my answer and yours) :)

Although your answer seems to be more comprehensible and better...
 

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