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Exactly one self-draw

CombinatoricsMedium

$5$ friends each write their name on a slip. The slips are shuffled and each friend draws one slip, with all $120$ assignments equally likely. What is the probability that exactly $1$ friend draws their own name?

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Hint

Choose who draws their own name, then require that the other friends all miss.

Solution

Choose the friend with the fixed point: $(1 5) = 5$ ways.
The other $4$ friends must all draw a wrong slip: $4! (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 9$ ways.
The number of favorable assignments is $5 \times 9 = 45$ .
The probability is $\frac{45}{120} = \frac{3}{8}$ .

Answer: $\frac{3}{8}$

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