The problem is written as follows:

Each of 10 employees brings one (distinct) present to an office party. Each present is given to a randomly selected employee by Santa (an employee can get more than 1 present). In how many ways can at least two employees get no presents?

The total number of ways to give out presents without restriction is: 10^{10}

I am going to count this using the complement, i.e 0 people recieve no presents or 1 person recieves no presents.

Case 1: 0 people receive no presents
Each person must get at least 1 present, so everyone gets exactly 1 present. Number of ways: 10!

Case 2: 1 person recieves no presents

First, we pick the person who gets no presents: \binom{10}{1}
Some other person must get two presents: \binom{9}{1}
Pick 2 of the 10 presents to give that person: \binom{10}{2}
Hand out the remaining 8 presents (one to each of the remaining 8): 8!

Answer for this case: \binom{10}{1}\binom{9}{1}\binom{10}{2}*8!

Overall answer: 10^{10}-[10! + \binom{10}{1}\binom{9}{1}\binom{10}{2}*8!]

The case 2 answer kind of bothers me since we flip flop between permutations and combinations.

Thank you for your help.