How many ways of arranging $7$ examinees are possible? 
In an examination hall, there are $20$ chairs in a row. In how many ways the seats of $7$ examinees can be arranged such that no two examinees can occupy two consecutive chairs?

I tried doing it by using the binary string method adn got the answer to be $\dfrac{14!}{7!}$, but the correct answer is the book is $8^7$. i just needed to verify the answer. If the correct answer is $8^7$, then please explain the solution.
My Approach:
Let each arrangement of states of chairs (empty or non empty) correspond to a binary string of length $20$.
For example, If chair no. $1,5,7,9,13,15,19$ are non empty while all other chairs are empty, then this must correspond to string $10001010100010100010$
Now for 7 examinees to sit, there would be $7$ non empty chairs and $13$ empty chairs. This means that the string will consist of $7$ ones and $13$ zeroes. Since no examinees must sit in consecutive chairs, no two $1s$ must appear consecutively in the string. Thus the string can be like this:
$$\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_$$
There are $14$ "gaps" where one can appear. We need to choose $7$ gaps out of $14$ for $1$ to appear at whereas the remaining gaps will be removed from the final string. This can happen in $\binom{14}{7}$ ways.
This gives us the arrangement of chairs. Now in any given arrangement, $7$ examinees can be permuted in $7!$ ways.
$\therefore$ The total number of ways of seating $7$ examinees in $20$ chairs such that no two examinees sit on consecutive chairs are $\binom{14}{7}\cdot 7!=\dfrac{14!}{7!}$
THANKS
 A: I did a brute force calculation, and it confirms that you are right.  I counted all the numbers from $1$ to $2^{20}$, whose binary representations have $7$ one-bits, no two of which are consecutive.  There are $3432$ such.  This gives the number of ways to choose the seats.  Then I multiplied by $7!$ to count the ways to seat the students.  The product turned out to be $$\frac{14!}{7!}$$ in agreement with your answer.
A: Let $x_1,...,x_6$ be the gaps (number of empty seats) between successive students.

Let $x_0$ be the number of empty seats to the left of the left-most student, and let $x_7$ be the number of empty seats to the right of the right-most student.

Then the number of legal choices for the seats to be used by the students is the number $S$ of $8$-tuples $(x_0,...,x_7)$ of integers satisfying the equation
$$
x_0+\cdots+x_7=13
$$
where $x_1,...,x_6$ are positive and $x_0,x_7$ are nonnegative.

Equivalently, letting 
$$
\;\;\;\;\,
\left\lbrace
\begin{align*}
y_0&=x_0+1\\[4pt]
y_k&=x_k\text{, for}\;1\le k\le 6\\[4pt]
y_7&=x_7+1\\[4pt]
\end{align*}
\right.
$$
$S$ is the number of $8$-tuples $(y_0,...,y_7)$ of positive integers satisfying the equation
$$
y_0+\cdots+y_7=15
$$
By the Stars-and-Bars method we get
$$
S=\binom{15-1}{8-1}=\binom{14}{7}
$$
hence, since for a given choice of seats for the students, there are $7!$ ways to seat the students, it follows that number of legal seatings is
$$
7!{\,\cdot\,}\binom{14}{7}=\frac{14!}{7!}
$$
matching your answer.

Thus, the answer key's answer of $8^7$ is not correct for the problem as stated.
