What is the chance all choices are different? Suppose everyone (7 people) chooses independently and randomly out of 7 choices.
What are the odds that no two people have picked the same option?
I feel like the answer is $5040/823543,$ but I am rusty with anything math related.

There are $7$ ways for one person to make a choice.  Assuming they choose uniformly, which means no person favors any choice over any other, each option therefore has a chance of $1/7.$  (This follows directly from the probability axioms, which assert the sum of all seven equal chances equals $1.$)
When $n$ people independently make choices, the very definition of independence means that any given array of choices has a chance of $1/7\times \cdots \times 1/7 = 1/7^n.$
An array consisting of all distinct choices denotes a permutation of those choices.  There are $7$ permutations of one choice, $7\times (7-1)$ permutations of two choices (because the second cannot agree with the first), $7\times (7-1)\times (7-2)$ permutations of three choices (two possibilities are excluded from the first choice), and so on.  The number of permutations of all $7$ options therefore is $7\times 6\times \cdots \times 1 = 7! = 5040.$
Since all permutations of the choices are distinct, the probability axioms tell us to add the chances of all these permutations.  This amounts to multiplying the common chance of $1/7^n$ by the number permutations, giving
$$\frac{1}{7^7} \times 7!  = \frac{5040}{823543} \approx 0.006119899$$
as stated in the question.
