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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 one present). What is the probability that at least two employees receive no presents?

I was thinking about tackling this as a, "find the inverse" type problem, where the inverse is: what is the probability that at least 9 employees receive at least 1 present. So I do stars and bars w/ 9 fixed stars... which is C(1 + 10 - 1, 1) which is nonsense. Help!

Note: the answer, which I don't understand, is 1−(10!−10×9×10!/2!)/10^(10).

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    Note that stars and bars tells you the number of outcomes but does not return equally likely outcomes. Take a smaller example, two people and two presents. Stars and bars will tell you there are three ways for the presents to be distributed, namely $[\star\star][~],~~[~][\star\star],$ and $[\star][\star]$. What it doesn't tell you is that $[\star][\star]$ is twice as likely to occur as the others. – 2017-02-03

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As each present can be given to $10$ employees independently, there are $10^{10}$ ways to give out the presents to the employees. Do you know which of the $10$s is the number of employees and which is the number of presents? You are correct that it is easier to count how many of these give presents to $9$ or $10$ employees. For $10$ you can just line the employees up, then each ordering of the presents in a row, of which there are $10!$, gives a present to each employee. For $9$, you have $10$ ways to pick the person who gets two, $10 \choose 2$ ways to select the two they get, then add an empty box to the remaining eight and distribute them to the $9$ remaining employees in $9!$ ways. So the number of ways to give the presents to $9$ or $10$ different employees is $10^{10}+10{10\choose 2}9!$

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    This makes sense, but I can't reconcile it with what this term should be according to the answer, which says it should be (10!−10×9×10!/2!). – 2017-02-03
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    The two agree. ${10 \choose 2}=\frac {10 \cdot 9}2$ and $10 \cdot 9!=10!$ – 2017-02-03