No of possible binary sequences with at most 2 consecutive zeroes

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.

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 $000$.