Giving presents to 8 employees out of 10 Each of 10 employees brings one (distinct) present to an office part. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents? How would you approach this?
 A: It is easier to find the probability that all get a present, or all but one get a present.
There are $10^{10}$ equally likely ways for the presents to be distributed. For we can line up the presents, and for each of the $10$, there are $10$ people it can go to.
There are $10!$ ways to distribute the presents so that everybody gets one.
We now count the ways to distribute so that exactly one person gets left out. The sad person can be chosen in $\binom{10}{1}$ ways. For each such choice, the lucky person who will get $2$ can be chosen in $\binom{9}{1}$ ways. The presents she gets can be chosen in $\binom{10}{2}$ ways. And the remaining $8$ presents can be distributed to the remaining $8$ people, one to each, in $8!$ ways. 
A: Total Number of ways 10 gifts could be distributed is 10^10
Number of ways to distribute the presents so that everybody gets one = 10P10 = 10!
Number of ways to distribute the presents so that 9 people get it, is the same as one 
person gets two gifts and the remainder of 8 of 9 get one each.  The two gifts that the 
lucky person gets can be chosen from 10.  Hence 10C2.  Clubbing those two gifts as one, now 
you have 9 presents to give to 10 people.  This could be done by 10P9 ( order matters).  
Thus for 9 people to get 10 presents the total number of ways = 10c2*10P9 = 45*10!
Thus the required probability = 1-P(10)-P(9) = 1 - [(10! + 45*10!)/10^10] = 
The problem becomes challenging once the wording changes to atleast (3,4,5...) do not get 
it. 
