What is the probability that $5$ men won't sit next to each other in a row of $20$ chairs? What is the probability that none of $5$ men won't sit next to each other in a row of $20$ chairs if only these $5$ men sit?
I have tried to solve this question. I got the total ways to be $\dbinom{20}{5}=15504$ and I got the total ways to be $1365$. This would make the answer $\dfrac{455}{5168}$. I'm not too sure if this is the right answer since my counting method was tedious. So mainly, I'm looking for a more efficient and accurate way. Thanks!
Any solutions?
 A: 
We decode empty chairs with $0$s and non-empty chairs with $1$s and consider binary strings of length $20$ with $15$ zeros and $5$ ones. Since no men are allowed to sit next to each other we are asking for words which do not contain a string $11$.
The number of all binary strings of length $20$ with $15$ zeros is
  \begin{align*}
\binom{20}{15}=15504
\end{align*}

In order to count the number of strings of length $20$ without having a substring $11$ we consider words
with no consecutive equal characters at all. These words are called Smirnov words or Carlitz words. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.) 

A generating function for the number of Smirnov words over a binary alphabet is given by
  \begin{align*}
\left(1-\frac{2z}{1+z}\right)^{-1}\tag{1}
\end{align*}

We replace occurrences of $0$ in a Smirnov word by one or more zeros since there are no restrictions to them. This corresponds to a substitution of
\begin{align*}
z\longrightarrow z+z^2+\cdots=\frac{z}{1-z}\tag{2}
\end{align*}
Since we want to look for strings of length $20$ with precisely $5$ ones, we mark them with $t$
\begin{align*}
z\longrightarrow tz\tag{3}
\end{align*}

We obtain by substituting (2) and (3) in (1) a generating function A(z,t)
  \begin{align*}
A(z,t)&=\left(1-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{tz}{1+tz}\right)^{-1}\\
&=\frac{1+tz}{1-z-tz^2}
\end{align*}
To obtain the number of words of length $20$ with $5$ ones we calculate with some help of Wolfram Alpha
  \begin{align*}
[z^{20}t^5]A(z,t)=4368
\end{align*}
  and conclude the probability that $5$ man won't sit next to each other in a row of $20$ chairs is
  \begin{align*}
\frac{4368}{15504}\simeq 0.2817
\end{align*}

A: Sometimes it's easier to count the number of ways to construct a desirable arrangement than to outright count the number of arrangements.  This is a method called "constructive counting."
Start with the five men A, B, C, D, E sitting in a row and count the number of ways to arrange the chairs.  There are 6 places we can place the 15 chairs -- before A, between A and B, between B and C, etc.  To make sure that none of the men are sitting next to each other, we first place one chair between A and B, one between B and C, etc.  Then we have 11 chairs to distribute among 6 different places -- which is easily counted with the "stars and bars" or "balls and urns" method.
A: Recall that the number of ways to divide $n$ into $k$ parts is the number of ways to arrange $n$ identical dots and $k-1$ identical separator bars, which is $\frac{(n+k-1 )!}{n! (k-1)!} = {n+k-1 \choose k-1}$.
Once the $5$ men are seated in some order, the remaining 15 chairs need to be distributed into 6 places (4 places are between them and two are in the left and right ends).  The number of ways to do this is the number of ways to divide $15$ into 6 parts, which is ${15+5 \choose 5} = 15504$, which is the denominator.
In how many ways can we distribute $15$ chairs into 6 places so that there is at least one chair between each consecutive pair of men?  First we place one chair into each of these 4 places.  The remaining $11$ chairs need to be distributed among the $6$ places. This can be done in ${11+5 \choose 5} = 4368$ ways.  This is the numerator, and the probability in question is therefore $4368/15504 \approx 0.2817$. 
