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CombinatoricsHard

$6$ guests each have a personal seat among $6$ numbered seats, and every guest takes exactly one seat, all different. How many seatings have exactly $2$ guests in their own personal seats?

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Hint

Choose who sits correctly, then count the seatings of the rest where nobody sits correctly.

Solution

Choose the $2$ guests who sit correctly in $(2 6) = 15$ ways.
The other $4$ guests must all sit in wrong seats, which is the number of derangements of $4$ objects.
That number is $4! - 4! + \frac{4 !}{2 !} - \frac{4 !}{3 !} + \frac{4 !}{4 !} = 9$ .
Multiplying gives $15 \times 9$ .
Hence the number of seatings is $135$ .

Answer: $135$

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