Formula to calculate the number of possible positions for $x$ numbers What formula do I use to calculate the number of possible positions for $x$ numbers?

Let's say I have $3$ people in a race. What are all the possible combinations of the order they can finish in? Let's assume ties are not possible. I heard I use factorial but its been a while since I have used factorials. So I want to verify.
It is easy to enumerate for three people. Lets call the people $x_1,x_2,x_3$. The possible orderings are shown below.


$x_1,x_2,x_3$
$x_1,x_3,x_2$
$x_2,x_1,x_3$
$x_2,x_3,x_1$
$x_3,x_1,x_2$
$x_3,x_2,x_1$


In general, in a $n$ people race, lets denote the people by $x_1,x_2,\ldots,x_n$. The first position can be taken by any one of the $n$ individuals. Hence, there are $n$ options for the first place. Now given that the first position is taken by $1$ individual, for each of the $n$ options for the first position, the second position can be taken by any one of the remaining $n-1$ individuals. Proceeding like this, in general the $k^{th}$ position can be taken by any one of the remaining $n-k+1$ individuals. Hence, the total number of ordering is given by $$n \times (n-1) \times \cdots \times (n-k+1) \times \cdots 2 \times 1$$which is nothing but $n!$.

In you case, $n=3$ gives us $3! = 6$ which matches with our enumeration.
Decent answers. However, with no duplicates, the solution may be more simply explained.

Since no ties are involved (duplicates) each position can have one less than the next highest finisher and thus we get:

3 * 2 * 1 = 6

If there were 5 people in the race, the solution would then be 5 * 4 * 3