Roll a fair, six-sided die 10 times. What is the probability that each side appears at least once?

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The answer that I got, which is incorrect, is the multinomial coefficient with $n=10$ and $k$ values $(1,1,1,1,1,5)$ multiplied by $6$ divided by the total number of outcomes, $6^{10}$. This in decimal form is equal to $0.003$.

I know the answer is $38045/139968$

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Actually there are more partitions, for example $(1,1,1,1,2,4)$ or $(1,1,2,2,2,2)$.

In order to consider all the possible partitions you have to evaluate the Stirling number of the second kind $$S(10,6)=22827$$ (see the table of values), that is the number of ways to partition a set of $10$ objects (the number of throws) into $6$ non-empty subsets (the number of values of a die). Then the required probability is equal to $$\frac{6!S(10,6)}{6^{10}}=\frac{38045}{139968}.$$