Probability each of six different numbers will appear at least once?

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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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 $$\frac{(6)\binom{7}{2}(5!)}{6^7}.$$
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