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If seven balanced dice are rolled, what is the probability that each of the six different numbers will appear at least once?

My solution is $\frac{6 \cdot 7!}{6^7}$.

Is this correct? How would I implement a solution using multinomial coefficients?

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up vote 6 down vote accepted

There are $6^7$ sequences of $7$ tosses, all equally likely.

We now count the sequences in which all the numbers appear. The number that appears twice can be chosen in $6$ ways. For every such choice, where it appears can be chosen in $\binom{7}{2}$ ways. And the rest of the numbers can be put in the empty slots in $5!$ ways. So our probability is $$\frac{(6)\binom{7}{2}(5!)}{6^7}.$$

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What is wrong with assigning the seven slots $7!$ ways and multiplying it by 6 to represent a number that can be 6 different values? – idealistikz Dec 24 '12 at 5:47
It double counts, for example, the sequence $1123456$. You might want to apply the same reasoning to a $2$-sided die (coin). Your procedure will give a probability greater than $1$. – André Nicolas Dec 24 '12 at 5:50
What if the question asks for the probability of every possible number will appear? – Tim Dec 16 '15 at 20:12
For this situation, that is another wording for the same question. – André Nicolas Dec 16 '15 at 20:45

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