There are $2^n$ possible sequences of heads and tails (where order matters). Of these, in exactly $\binom{n}{k}$ of them you have $k$ heads (pick in which positions you have heads, the remaining positions will be tails).
There are, however, only $n+1$ possible total outcomes with $n$ coin tosses, if all you care about is how many heads and how many tails you got; namely, you can have $0$ heads, $1$ head, $2$ heads, $3$ heads, and so on until you get to $n$ heads.
If you don't care about order and are only interested in the possible ratio of heads to tosses, the possibilities are $0$, $\frac{1}{n}$, $\frac{2}{n}$, $\frac{3}{n},\ldots,\frac{n-1}{n}$, and $1$.
If you are interested in order, and want to know the probability that a sequence of $n$ tosses will result in exactly $k$ heads, that number is $\frac{\binom{n}{k}}{2^n}.$ So the probabilities would be:
- Of getting no heads, $\displaystyle\frac{1}{2^n}$.
- Of getting exactly one head, $\displaystyle \frac{\binom{n}{1}}{2^n} = \frac{n}{2^n}$.
- Of getting exactly two heads $\displaystyle\frac{\binom{n}{2}}{2^n} = \frac{n(n-1)}{2^{n+1}}$.
- Of getting exactly three heads, $\displaystyle\frac{\binom{n}{3}}{2^n} = \frac{n(n-1)(n-2)}{2^n 3!}$.
etc. The odds of getting exactly $k$ heads are the same as the odds of getting exactly $n-k$ heads.
It is the latter that I would consider a "frequency", as $\binom{n}{k}/2^n$ is how frequently you would expect to get a set of $n$ tosses that have exactly $k$ heads in it.