combinatorics: 5 people picking 10 seats when there must be at least one space between them I have this question: How many seating arrangements are there for $5$ people to sit in $10$ seats in a row, when $2$ people can't sit next to each other?
My idea:
If there must be at least one space between every 2 people, the spaces must be something like this:
_ _ s _ s _ s _ s _
There must be $2$ open seats next to each other, so they have $5$ options of where to be, and the person who sits there, have $2$ options to choose from.
The seating arrangements for $5$ people is $5!$ in a standard row,
so overall: $5 \cdot 2 \cdot 5! = 1200$.
My answer is wrong, so I was wondering what is a better way to think about it.
 A: Hand each person a chair.  Place the other five seats in a row, leaving spaces between each empty chair and at the ends of the row.  This creates six spaces in which the people can place their chairs.
$$\square c \square c \square c \square c \square c \square$$
To ensure that no two people sit in adjacent seats, the people must choose five of these six spaces in which to place a chair, which can be done in $\binom{6}{5}$ ways.  They can arrange themselves in the five selected spaces in $5!$ ways.  Hence, there are
$$\binom{6}{5}5!$$
seating arrangements in which five people can be seated in $10$ chairs so that no two of the people are adjacent.
A: 
How many seating arrangements are there for $5$ people to sit in $10$ seats in a row when no $2$ people can sit next to each other?

$\underbrace{p\_p\_p\_p\_p}_{9}$
Put the last seat in one of the $6$ places and arrange $5$ persons. Answer $=\boxed{6\cdot5!}$.
A: You must have p_p_p_p_p and one more empty seat somewhere. There are $6$ possible locations for the remaining empty seat: on one end, or between two people. The $5$ people can be arranged in $5!$ different ways in their chosen seats, so altogether there are $6\cdot5!=720$ arrangements.
Notice that you don’t have to have two empty seats next to each other: you can have, for instance, _p_p_p_p_p. If you want to use that line of argument, you should notice that there are just $4$ places between two people, so there are $4$ places to put the pair of seats. But then there are also the arrangements _p_p_p_p_p and p_p_p_p_p_, for a total of $6$.
