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

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.
