combinatorics: 5 people picking 10 seats when there must be at least one space between them

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 $6$.


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.


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 $=\boxed{6\cdot5!}$.

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Combinatorics

Permutations

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