6. During a party, three girls and three boys sit down randomly at a round table. What is the probability that no two people of the same gender are sitting next to each other?
(A) $\frac{1}{6}$
(B) $\frac{1}{10}$
(C) $\frac{3}{20}$
(D) $\frac{1}{12}$
(E) $\frac{11}{36}$.
6. The answer is (B). If we fix a boy, there are 5! possible permutations of the other 5 guests, of which those in which boys and girls alternate are $3 \cdot 2 \cdot 2$ because the person to the right of the fixed boy can be any of the 3 girls, the person further to the right can be one of the two remaining boys, then there must be one of the other two girls, and the last two positions must necessarily be occupied by the last boy and the last girl.

The solution is therefore the number of favorable cases divided by the number of possible cases, that is $\frac{3 \cdot 2 \cdot 2}{5 !}=\frac{1}{10}$.