Tanya prepared 4 different letters to 4 different addresses. For each
letter, she prepared one envelope with its correct address. If the 4
letters are to be put into the four envelopes at random, what is the
probability that only one letter will be put into the envelope with
its correct address?
I have solved in this way:
The way of assigning in is = 4! = 24
If we think only one assigning will be correct then we can assign this in $$ * \times 3 \times 2 \times 1 =6 \ \ ways $$
then the probability would be 6/24 = 1/4.
But the answer is given that, 1/3.
Best Answer
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.