Question

n objects are distributed at random among n persons. The number of ways in which this can be done so that at least one of them will not get any object is

• A. $n!-n$
• B. $n^n-n$
• C. $n^n-n^2$
• D. $n^n-n!$

Answer 1

Answer: (D)

$n^n-n!$

Answer 2

The initial step involves distributing the n objects among n persons. The first object can be given to any of the n persons, and similarly, the subsequent objects can be distributed to any of the n persons independently. Thus, the total number of ways to distribute the objects randomly among the n persons is n^n.

Now, we're interested in the probability that each person gets exactly one object. There are n! (n factorial) ways to distribute the objects such that each person gets one object. This is because there are n choices for the first object, (n - 1) choices for the second object (since it can't go to the person who got the first object), (n - 2) choices for the third object (since it can't go to the first two persons), and so on until the last person, who has only one choice.

Therefore, the probability of each person getting exactly one object is n! / n^n.

However, we want to find the probability that at least one person does not get any objects. This is the complement of the earlier probability. So, the probability that at least one person doesn't get any objects is 1 minus the probability that each person gets exactly one object:

1 - (n! / n^n) = (n^n - n!) / n^n.

This is the final probability that at least one person does not get any objects.

The correct answer is option (D): $n^n-n!$