Seven balls labeled with number 1 to 7 are distributed randomly among seven boxes also labeled with number 1 to 7. And each box can contain at most 1 ball. What is the probability that exactly 2 balls go to boxes with the same number?
It seems that we should fix the 2 correct boxes first, and consider the remaining 5 boxes with inclusion-exclusion principle. But I am not sure how to apply the principle in this situation, what event we should denote here. Can anyone explain it for me?
You're right about the first step, however I disagree with your other steps.
First, we must choose and place two balls in their corresponding boxes. This can be done in ${7 \choose 2}$ ways.
Next, we need to place the remaining five balls in such a way that none of them are in their corresponding boxes. The number of ways to do this is equal to the number of derangements of $5$. So, this can be done in $44$ ways.
To find the probability of this situation occurring, we note that the balls can be arranged in a total of $7!$ ways. Therefore, the probability is
$${7 \choose 2}\frac{44}{7!}=\frac{11}{60}.$$