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

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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...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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