No of possible binary sequences with at most 2 consecutive zeroes The problem:
How many such sequences of ten $1's$ and ten $0's$ are possible such that they don't contain three or more consecutive zeroes? (Eg. $00101001001101110011$)
I know there are $20!/10!10!$ total sequences. How do I filter out the ones which have more than two consecutive zeroes somewhere?
 A: The answer can be written as
$${11\choose0}{11\choose10}+{11\choose1}{10\choose8}+{11\choose2}{9\choose6}+{11\choose3}{8\choose4}+{11\choose4}{7\choose2}+{11\choose5}{6\choose0}$$
That is, imagine you've got a string of $10$ ones and you now need to insert $10$ zeroes in doublets and singlets. There are $11$ places where the doublets and singlets can go: $9$ places between ones and $2$ places at the far left and far right.  If you first insert $k$ doublets, you'll be left with $10-2k$ singlets to insert, with $11-k$ places to insert them, for ${11\choose k}{11-k\choose10-2k}$ choices in all. The number of doublets can range from $0$ to $5$, hence the sum.
Remark: The expression calculates out to $24{,}068$, which is is divisible by $547$, as the OP says (in a comment below true blue anil's answer) must be the case.
A: This answer is based upon the Goulden-Jackson Cluster Method.

We consider the words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{0,1\}$$ and the set $B=\{000\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coefficient of $s^n$ being  the number of searched words of length $n$.

According to the paper (p.7) the generating function $f(s)$  is
\begin{align*}
f(s)=\frac{1}{1-ds-\text{w}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{V}|=2$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{w}(\mathcal{C})=\text{w}(\mathcal{C}[000])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{w}(\mathcal{C}[000])&=-(sz)^3-sz\cdot \text{w}(\mathcal{C}[000])-(sz)^2\cdot\text{w}(\mathcal{C}[000])\tag{2}\\
\end{align*}
where we additionally mark occurrences of zeros with $z$. We obtain
\begin{align*}
\text{w}(\mathcal{C})=\text{w}(\mathcal{C}[000])=-\frac{(sz)^3(1-(sz))}{1-(sz)^3}
\end{align*}

It follows from (1) and (2)
\begin{align*}
f(s)&=\frac{1}{1-ds-\text{w}(\mathcal{C})}\\
&=\frac{1}{1-(1+z)s+\frac{(sz)^3(1-sz)}{1-(sz)^3}}\\
&=\frac{1+sz+s^2z^2}{1-s-zs^2-z^2s^3}\\
&=1 + (z+1)s + (z+1)^2 s^2 + (3z^2+3z+1)s^3 +\cdots\\
&\qquad+ \left(z^{14}+112z^{13}+\cdots+\color{blue}{24\,068}z^{10}+\cdots+20z+1\right)s^{20}+\cdots\\
\end{align*}
The last line was calculated with some help of Wolfram Alpha. The coefficient of $z^{10}s^{20}$  shows that out of $2^{20}=1\,048\,576$ binary words of length $20$ from the alphabet $\{0,1\}$ there are $\color{blue}{24\,068}$  words containing ten $0$s and ten $1$s which do not contain $000$.

A: The ten $1's$ can be looked upon as $10$ bars forming $11$ "compartments" in which a maximum of two $1's$ can be put.
This can be solved using stars and bars  with inclusion-exclusion
$N = \binom{20}{10} - \binom{11}1\binom{17}{10} +\binom{11}2\binom{14}{10} - \binom{11}3\binom{11}{10} = 24068$
