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Keeping $2$ people apart

CombinatoricsMedium

$6$ distinct people sit in a row of $6$ chairs. $2$ specific people must not sit next to each other. How many seating orders are possible?

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Hint

Subtract the bad seatings from all seatings.

Solution

Without any restriction there are $6! = 720$ seatings.
To count the bad seatings, glue the $2$ specific people into one block, which leaves $5$ units.
The units can be ordered in $5! = 120$ ways and the pair can swap in $2$ ways, so there are $240$ bad seatings.
Subtracting gives $720 - 240$ .
Hence the number of valid seatings is $480$ .

Answer: $480$