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?

Firstly, there are $10^{10}$ total ways to give the $10$ employees the $10$ presents. So this is our denominator.

My attempt was to consider the complement and consider the number of ways that either $0$ employees receive no presents (every employee gets a present) or $1$ employee receives no present.

Case 1: $0$ employees

There are $10$ employees and $10$ presents. So there are $10^{10}$ ways to give the presents.

Case 2: $1$ employee

Step 1: Decide which employee receives no presents: $10$ possibilities.

Step 2: Distribute the $10$ presents to the remaining $9$ employees: $9^{10}$ ways.

So the number of ways in which at least $2$ employees receive no presents is: $1-(10^{10}+9^{10}$).

So my final answer is: $1-\displaystyle\frac{(10^{10}+9^{10})}{10^{10}}$.

However, this answer does not match the answer in my textbook. Which is: $1-\displaystyle\frac{10!-10\times 9 \times \frac{10!}{2!}}{10^{10}}$

Where did my attempt go wrong and how can I correct it?

  • $\begingroup$ How many ways can you distribute gifts to 8 employees $\endgroup$ Oct 8 '18 at 0:12
  • $\begingroup$ $8^{10}$ ways? @RushabhMehta $\endgroup$
    – rover2
    Oct 8 '18 at 0:14
  • $\begingroup$ $10! - 10\cdot 9\cdot \frac{10!}{2!}$ is a negative value which makes the textbook answer greater than $1$. $\endgroup$
    – Phil H
    Oct 8 '18 at 0:39
  • $\begingroup$ @PhilH i am not overlooking the fact that the answer in the textbook can be wrong...this textbook is known for having errors. I am just trying to understand how to complete the problem CORRECTLY, rather than understanding the answer in the textbook. $\endgroup$
    – rover2
    Oct 8 '18 at 0:49
  • $\begingroup$ @rover2 that's your answer $\endgroup$ Oct 8 '18 at 1:53

Where P(x) is the probability that x number of people do not receive a gift, then

$P(\ge 2) = 1 - (P(0)+P(1))$

There are $10!$ ways that all $10$ people can get one gift and $10^{10}$ ways to distribute $10$ gifts to $10$ people.

$$P(0) = \frac{10!}{10^{10}}$$

Then we have $10$ ways for someone not to get a gift and for each one of those there are $9$ ways for someone to get $2$ gifts and for each of those $2$ people combinations there are $\frac{10!}{2!}$ ways to combine them with $8$ other people who each receive one gift.

$$P(1) = \frac{9\cdot 10\cdot 10!}{10^{10}\cdot 2!}$$

$$P(\ge 2) = 1 - (\frac{10!}{10^{10}}+\frac{9\cdot 10\cdot 10!}{10^{10}\cdot 2!}) = .98330752$$


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