In a café, at three two-person tables, a total of six people take seats. Among them, three drink coffee and three drink tea. What is the probability that there is a table where both people drink tea?
Solution. The 3 coffees and 3 teas can be served to the 6 guests in

$$
P_{6}^{3,3}=\frac{6!}{3!3!}=20
$$

ways.

If there is a table where both guests drink tea, then among the other 4 guests, 3 coffees and 1 tea need to be distributed, which can be done in

$$
P_{4}^{3,1}=\frac{4!}{3!1!}=4
$$

ways. Since there are 3 tables, this can happen at any of them, so the number of favorable cases is $3 \cdot 4=12$.

The probability we are looking for is the ratio of the number of favorable cases to the total number of cases, $P=\frac{12}{20}=0.6$.