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A secretary writes letters to 8 different people and addresses 8 envelopes with the people's addresses. He randomly puts the letters in the envelopes. What is the probability that he gets exactly 6 letters in the correct envelopes?

I made a start by finding the total outcomes, and then subtracting the probability that he got two letters wrong, but then hit a dead end.

Could someone please give me a solution to this problem, or maybe a hint?

Thanks.

coolcheetah
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    Look at "derangements" from wikipedia. – voldemort Jan 27 '15 at 17:20
  • Hint: How many ways are there to choose the $6$ lucky people who get the right letter? – André Nicolas Jan 27 '15 at 17:25
  • This question is easily solved by looking for derangements and noting that the number of permutations is $8!$. Apply the answer in the duplicate to get a probability by dividing number of favourable results by total possible. – AlexR Jan 27 '15 at 17:25
  • @AlexR: Because of the choice of numbers, the current problem is much simpler than the linked problem. – André Nicolas Jan 27 '15 at 17:27
  • "First $K$ are deranged" does not involve the binomial coefficient, and requires more background. I can see the current question as being asked very early in the introduction to counting. There have been closer "abstract duplicates" where one asks for the probability that exactly $m$ letters go to the right person. – André Nicolas Jan 27 '15 at 17:53

2 Answers2

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If exactly 6 of 8 are right, then exactly 2 of them have been swapped. So choose two of then to exchange. $\binom{8}{2} = 28$.

NoName
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You can think of it like this:

The envelopes have fixed positions $1, 2, 3, 4, 5, 6, 7, 8$. Now which letter gets which envelope is described by an 8-tuple. The 8-tuple $(2, 1, 3, 4, 5, 6, 7, 8)$ means that he gets every letter right, except for letter 1 and 2.

Your sample space is the set of all permutation of the number 1, ... 8:

$$\Omega = \{(x_1, \dots, x_8) | x_i \in \{1, \dots, 8\}, i \neq j \Rightarrow x_i \neq x_j\}$$

$$|\Omega| = 8! = 40320$$

How many ways are there to get exactly 6 right? 6 out of 8...

$$\binom{8}{6} = \frac{8!}{6! (8-6)!} = \frac{7 \cdot 8}{2} = 7 \cdot 4 = 28$$

This makes

$$P(\text{get exactly 6 letters right}) = \frac{28}{40320} = \frac{1}{1440} \approx 0.06944 \%$$

Martin Thoma
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