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Dinner place cards

CombinatoricsMedium

$6$ guests at a dinner each have a personal place card. The cards are shuffled and each guest takes one card. How many of the $720$ assignments have exactly $2$ guests holding their own card?

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Hint

Pick the $2$ lucky guests, then count assignments of the other guests with no match.

Solution

Choose the $2$ guests who hold their own cards: $(2 6) = 15$ ways.
The remaining $4$ guests must all hold wrong cards: $4! (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 9$ ways.
Multiply the independent counts: $15 \times 9 = 135$ .
The answer is $135$ .

Answer: $135$

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