Choosing $6$ of $20$ people in a row such that there is at least one other chosen beside any of them There are $20$ people in a row. What is the number of ways to choose $6$ of them such that there is at least one adjoining person beside every chosen person.
I was trying to count number of ways of its complement. But it gets harder. Actually, I have no idea how to find an approach. 
 A: Each admissible arrangement can be encoded as a binary word of length $20$ containing exactly $6$ ones. The $14$ not chosen people (zeros) create $15$ slots where groups of chosen people (ones) may be squeezed in. These groups may have sizes $(6)$, $(4,2)$, $(3,3)$, and $(2,2,2)$. It follows that the total number of admissible arrangements is given by
$${15\choose 1}+2\cdot{15\choose 2}+{15\choose 2}+{15\choose3}=785\ .$$
A: Just to provide an alternative approch, we may utilise regular expressions.
This is equivalent to counting binary words of using the alphabet $\{A,B\}$ length $20$ with $6$ $A$s and 14 $B$s such that there are no isolated $A$s.
Consider that the regular expression for binary words with no isolated $A$s must be of the form
$$f=1 +f_A+f_B\tag{1}$$
where $1$ is the empty word, $f_A$ are all such words ending with $A$ and $f_B$ all such words ending with $B$. 
It should be clear that words in $f_A$ either consist entirely of $A$s or are $f_B$ with $2,3,4\ldots$ $A$s appended
$$f_A=(A+AA+AAA+\cdots) +f_B(AA+AAA+AAAA+\cdots)$$
$$\implies f_A=(1+f_B)A^2(1-A)^{-1}\tag{2}$$
similarly words in $f_B$ consist entirely of words containing $B$ or are $f_A$ with $1,2,3,\ldots$ $B$s appended
$$f_B=(1+f_A)B(1-B)^{-1}\tag{3}$$
solving $(2)$ and $(3)$ simultaneously gives
$$f_A=\left(\frac{BA^2}{(1-A)(1-B)}+\frac{A^2}{1-A}\right)\left(1-\frac{BA^2}{(1-B)(1-A)}\right)^{-1}\tag{4}$$
$$f_B=\left(\frac{BA^2}{(1-A)(1-B)}+\frac{B}{1-B}\right)\left(1-\frac{BA^2}{(1-B)(1-A)}\right)^{-1}\tag{5}$$
plugging $(4)$ and $(5)$ into $(1)$ yields (after some simplification)
$$f=\frac{A^2-A+1}{1-A-B+AB-A^2B}\tag{6}$$
For this problem we evaluate the coefficient of $A^6B^{14}$ in the expansion of $f$ using sage input
y=var('y')
taylor((x^2-x+1)/(1-x-y+x*y-x^2*y),(x,0),
(y,0),25).coefficient(x^6).coefficient(y^14)

This evaluates to
$$[A^6B^{14}]\frac{A^2-A+1}{1-A-B+AB-A^2B}=785\tag{Answer}$$
