## 40. At Lunch

Ten married couples are having lunch together. Five people are chosen by lot to set the table. What is the probability that among these five people there will not be a single married couple?
40. The number of ways to choose 5 people out of 20 is

$$
C_{20}^{5}=\frac{20!}{5!15!}
$$

The number of ways to choose such that no pair is represented in its entirety is

$$
C_{10}^{5} \cdot 2^{5}=25 \cdot \frac{10!}{5!5!}
$$

where $C_{10}^{5}$ is the number of ways to choose 5 pairs out of 10, and $2^{5}$ is the number of ways to choose one person from each of the 5 pairs.

The desired probability, therefore, is

$$
\frac{C_{10}^{5} \cdot 2^{5}}{C_{20}^{5}}=\frac{6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}{16 \cdot 17 \cdot 18 \cdot 19 \cdot 20} \cdot 2^{5}=\frac{168}{323}=0.52
$$