Let us count the number of ways that letter A can be in its correct envelope, which we also call A, while none of the others is in a correct envelope.
So B must go to C or D ($2$ choices).
If B goes to C, then C cannot go to B, else D would have to go to D. So C must go to D, and the fate of D is determined. So there is only $1$ way that B can go to C.
The same is true if B goes to D. So overall there are $2$ choices where A is the "fixed point."
That gives a total of $8$ possibilities. Divide by $4!$.
Remark: We could also solve this in a much more general way. We have $n$ letters, and want to find the probability that exactly $k$ of them end up in the correct envelope. The $n$ letters can be permuted in $n!$ ways. As in the answer above, that will be the denominator.
For the numerator, the lucky letters that end up in the right envelope can be chosen in $\binom{n}{k}$ ways. The unlucky $n-k$ letters can be all put into wrong envelopes in $D(n,k)$ ways, where $D(n,k)$ is the derangement number. For more about counting Derangements, please see Wikipedia. It uses the notation $!w$ for the number of derangements of $w$ objects.
For $3$ objects this is fairly trivial, as you can simply inspect each combination:
- $123:$ all letters are in the correct envelope
- $132:$ letter #$1$ is in the correct envelope
- $213:$ letter #$3$ is in the correct envelope
- $231:$ no letter is in the correct envelope
- $312:$ no letter is in the correct envelope
- $321:$ letter #$2$ is in the correct envelope
In $4$ out of $6$ combinations, there is at least one letter in the correct envelope.
Hence the probability of having at least one letter in the correct envelope is $\dfrac{4}{6}$.
Best Answer
The formula for the derangement can be used directly. The number of derangements is $$n! \left(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \ldots + \frac{(-1)^n}{n!}\right)$$ put the value of n and get your answer. I guess this is the fastest and safest way. Hope it helpful. Cheers.