Well, the probability of having at least $1$ person with the birthday
of $q$ is $n/365$.
Unfortunately, this isn't quite true. If you could guarantee that the $n$ people didn't share a birthday, then it would be true, but if you could guarantee that, there wouldn't be much of a birthday question! The actual probability that at least one person out of $n$ has $q$ as his or her birthday is
$$
1-\left(\frac{364}{365}\right)^n
$$
where we assume (as is typical for initial analyses of this problem) that birthdays are uniformly distributed over a $365$-day year. If $n$ is small, this is approximately equal to $n/365$ (and it is exactly true for $n = 1$), but this is increasingly untrue as $n$ becomes larger.
Note that if $n > 365$ then $n/365 > 1$, but we know it is perfectly possible for each of (let's say) $400$ people to happen not to have May $21$ for their birthday. The above expression gives that probability (usual assumptions applying).
To get the probability of $2$ people, you would multiply the
probability of having one person by $(n−1)/365$.
This also isn't true. In order to determine the probability that at least two people have $q$ as their birthday, it's simplest to subtract from the above probability the chance that exactly one person has $q$ as their birthday. This yields
$$
1-\left(\frac{364}{365}\right)^n
-n\left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{n-1}
$$
As before, what you say is approximately true for small $n$ (and exactly true for $n = 2$), but still is increasingly untrue for larger $n$, for nearly the same reason.
And if you wanted the probability of at least $k$ of the $n$ people sharing $q$ as their birthday, we could continue subtracting all the way up to $k-1$ people sharing $q$ as their birthday:
$$
1-\sum_{i=0}^{k-1} \binom{n}{i} \left(\frac{1}{365}\right)^i \left(\frac{364}{365}\right)^{n-i}
$$
So we can say, given $n$ people, the probability that $2$ of them will
both have the birthday $q$ (and therefore the same birthday) is $(n^2-n)/365^2$.
Even if the above were true, the analysis would have to be expanded to cover all possible $q$, rather than only a single $q$. If you evaluate my second expression above at $n = 2$, the result is $1/365^2$, the probability that two people share $q$ as their birthday. If you multiply that by $365$, the number of possible choices for $q$, you get $1/365$, which is the probability that two people share any day as their birthday. However, that expression cannot be extended to expand beyond $n = 2$ exactly, because it neglects the possibility that more than two people share the same birthday.
The standard analysis of the birthday problem is the most straightforward, in which the probability of at least two of $n$ people sharing a birthday is
$$
1-\left(\frac{365}{365}\right)
\left(\frac{364}{365}\right)
\left(\frac{363}{365}\right) \cdots
\left(\frac{366-n}{365}\right)
$$
Best Answer
The number of ways $n$ distinct dates can be selected from $365$ is $\binom{365}{n}$. After you have chosen you can arrange them in $n!$ ways. So total is $n!\times\binom{365}{n}$.
All $n$ people can have birthdays in $365$ possible days so the total ways is $365^n$ hence you get $$\frac{n!\binom{365}{n}}{365^n}$$
The formula you have in the denominator is different from what you intend, it is distributing 365 days among n people. (link) The formula you are using assumes that days and people aren't distinct from one another.
An equivalent number of possible 3 digit binary numbers. In this case the places are all distinct from one another and so are the digits. Hence the formula you use is based on an assumption not relevant to this problem.