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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.

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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:

$\frac{365!}{(365-n)!365^n}$

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  • $\begingroup$ I thought I would divide both enumerations by $n!$ since the order is not important... $\endgroup$ – bibo_extreme Dec 6 '15 at 16:27

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