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.
 A: 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.
A: The answer by whuber shows a nice way to approach this from first principles in combinatorics.  In this answer I will just note that this question is essentially asking for a probability from the classical occupancy distribution discussed in O'Neill (2020).  Given $n$ balls randomly allocated to $m$ bins, the probability mass function for the number of occupied bins $K_n$ is:
$$\mathbb{P}(K_n=k) = \text{Occ}(k|n,m) \equiv \frac{(m)_k \cdot S(n,k)}{m^n}.$$
In your case you have $n=7$ balls randomly allocated to $m=7$ bins, and your probability of interest is for full occupancy, which is:
$$\text{Occ}(7|7,7) = \frac{(7)_7 \cdot S(7,7)}{7^7} = \frac{7!}{7^7} = 0.006119899.$$
You can obtain values from the occupancy distribution in R using the probability functions in the occupancy package.  The present probability is easily obtained using the mass function for the occupancy distribution, using the query occupancy::docc(7, size = 7, space = 7).
Given your interest in determining the probability of full occupancy in this problem, you might be interested in the more general analysis of the occupancy number that is contained in the linked paper.  The occupancy distribution is quite interesting an it solves an antique problem in probability theory.  It also has a number of interesting generalisations.
A: I agree with your solution and @whuber's elegant solution, here is my rational to arrive at the same solution by simply counting all options first, then taking the ratio at the end to compute the probability:

*

*Let's assume the set of possible choices is:
$ O = \{1,2,3,4,5,6,7\} $


*We can imagine 7 people picking numbers from this set $ O $ generates a sequence of these 7 numbers, for example 2547551 would be one of many options. All possible ways of arranging 7 numbers in a sequence where repetition is allowed is: $ 7^7 $. We allow repetition here because the problem states that people choose independently, so one person's choice is not going to affect other person's choice.


*The only way that no 2 person pick the same option would be all sequences where no repetition is allowed. One possible option would be 1234567. Number of possible arrangements of 7 numbers with no repetition is 7!


*So the probability that no 2 people picked the same number is:
$ \frac{7!}{7^7} $
You can also run a simulation and arrive at the same solution if you have doubts.  I used Python.

*

*Generated all possible permutations of these 7 numbers, counted them, assigned the result to a.

*Counted the options where no number in a sequence was repeated, assigned the result to b.

*Computed b/a

Here is the Python program and its output in case anyone interested:
import itertools

O = [1,2,3,4,5,6,7]
a = 0
b = 0
all_combinations = itertools.product(O, repeat=len(O))
for sequence in all_combinations:    
    a +=1
    if len(set(''.join(map(str,sequence)))) == len(O):
        b += 1  
    
print(f"Number of all possible ways of arranging 7 numbers with repetition = {a}")
print(f"Number of all possible ways of arranging 7 numbers without repetition = {b}")
print(f"Probability of having no 2 people selecting the same option = {b/a}")

>>> Number of all possible ways of arranging 7 numbers with repetition = 823543
>>> Number of all possible ways of arranging 7 numbers without repetition = 5040
>>> Probability of having no 2 people selecting the same option = 0.006119899021666143

