What's the probability of getting $5$ different numbers but not any $6$ when throwing $5$ dice?

Question

I have $5$ dice, I throw them at once. What is the probability of getting $5$ unique numbers, i.e., $1\ \ \&\ \ 2\ \ \&\ \ 3\ \ \&\ \ 4\ \ \&\ \ 5$ in any order BUT NOT any $6$?

Of course they can be in any order as long as all $5$ dice are all unique numbers but not any $6$. I presume number of possibilities is divided by $6$ to the power of $5$ ($7776$)? But I do not know the number of possible permutations for $5$ unique numbers, excluding a $6$. Please help.

The question has an actual usefulness for me.

What is the probability of getting $5$ different numbers but not any $6$ when throwing $5$ dice at once?

Answer 1

Community wiki answer so the question can be marked as answered:

As noted in the comments, there are $5!=120$ different permutations of $5$ numbers, and as you wrote, there are $6^5=7776$ equiprobable outcomes, so the probability is

$$ \frac{120}{7776}=\frac5{324}\approx1.54\%\;. $$

Answer 2

You can have $5!$ cases for the $5$ different numbers, and clearly, you have $6^5$ cases as the total ones. So, the answer is $\color{red}{\frac{5!}{6^5}}$.