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.
On
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
$$P(\text{get exactly 6 letters right}) = \frac{28}{40320} = \frac{1}{1440} \approx 0.06944 \%$$
If exactly 6 of 8 are right, then exactly 2 of them have been swapped. So choose two of then to exchange. $\binom{8}{2} = 28$.