Question Consider $n$ people who are attending a party. We assume that every person has an equal probability of being born on every day of the year, independent of everyone else. Assuming that nobody is born on the $29^\mathrm{th}$ February and that $n\leq365$, find the probability that each person has a distinct birthday
Solution Enumerate the possible birthdays by $1, 2,\ldots,365$. Clearly this question allows repetition and does not distinguish between different orderings. We start by counting how many different ways there can be no two same birthdays. The number is:
$\dfrac{365\times364\times\ldots\times(365-n+1)}{n!}=\dfrac{365!}{n!(365-n)!}=\,{365\choose n}$
Now, the total number of different possibilities is $\dfrac{365^n}{n!}$.
So the probability is
$\dfrac{365!}{n!(365-n)!}\div\dfrac{365^n}{n!}$ etc.
I don't see why you divide by $n!$ when counting, especially the $\frac{365^n}{n!}$, remember this will not be an integer most of the time. Although your final result looks good.
Here is what I thought of:
The number of ways in which the birthdays can occur without repetition is $365\times \dots \times(365-n+1)!=\frac{365!}{(365-n)!}$. (Since there are $365$ options for person $1$, $364$ for person $2$ etc)
The number of ways in which the birtdays can occur with repetition is $(365)^n$. (Since there are $365$ options for every person)
Therefore the probability a birthday is repeated is:
$\frac{365!}{(365-n)!365^n}$