# Random Gift Giving at a Party - Combinatorics Problem

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?

• How many ways can you distribute gifts to 8 employees Oct 8 '18 at 0:12
• $8^{10}$ ways? @RushabhMehta Oct 8 '18 at 0:14
• $10! - 10\cdot 9\cdot \frac{10!}{2!}$ is a negative value which makes the textbook answer greater than $1$. Oct 8 '18 at 0:39
• @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. Oct 8 '18 at 0:49

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.

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

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