Counting strategies Exam Question How many different ways can people finish in 
i) a $4$ person race, ii) a $6$ person race,  iii) a $10$ person race
What I did:
$4^4 = 256$
$6^6 = 46,656$
$10^{10}$
as there are $4$ people and therefore 4th place 3rd place 2nd place and 1st place so $4$ to the power of $4$.
I don't know where I went wrong and is there a formulae for working out these type of questions?
Thank You and Help is Appreciated
 A: Just because there are $4$  ways to choose the first person and $4$ ways to choose the second person doesn't mean there are $16$ ways to choose both; some combinations such as $AAAA$ are't allowed (but are still counted).
Instead, try looking at the factorial function.
A: Approach with multiplication principle.


*

*Pick who the person who finishes first is

*Pick who the person who finishes second is.  (Note that the person who finishes first will not also finish second and so they should not be considered eligible anymore for any of the later positions)

*Pick who the person who finishes third is.  (Similarly, neither the person who finishes first nor the person who finishes second are able to also finish third, so there are two fewer people available to choose from here)

*$\vdots$
Multiplying the number of options available for each step gives the total.  Note that multiplication principle works so long as the number of options available at each step doesn't depend on the choices made in previous steps, but what those specific options are can indeed change based on previously made choices, such as here.
