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? 
 A: 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
$$\frac{(6)\binom{7}{2}(5!)}{6^7}.$$
A: First the all tossing outcomes is $6^{7}$
Then we consider choose 6 tosses from 7 tosses, there are 
$\left(\begin{array}{c}7\\6\end{array}\right)$ ways, then assign them outcome value 1, 2, 3, 4, 5, 6 in order satisfying the conditions at least appear once. there are 6! ways. Then for the remaining last toss we can choose any value, there are 6 ways. Mutilplying them together result in $\left(\begin{array}{c}7\\6\end{array}\right)\times6!\times6$. However it's not the final result!!!
Consider the situation 1234561, in my ways of counting, '1' has been repeatedly counting 2! times. Therefore we have to divide total answer by 2!.Hence there  are 
$\frac{\left(\begin{array}{c}
7\\
6
\end{array}\right)\times6!\times6}{2!}$ ways.
So our probability is $\frac{\left(\begin{array}{c}
7\\
6
\end{array}\right)\times6!\times6}{2!\times6^{7}}$
