10. Four mathematicians, two physicists, one chemist and one biologist take part in a table tennis tournament. The eight players are to compete in four pairs by drawing lots. What is the probability that no two mathematicians play against each other?
(1 mark)
Four mathematicians, two physicists, one chemist, and one biologist participate in a table tennis tournament. The eight players are to be divided into four pairs by drawing lots. What is the probability that no two mathematicians play against each other?
10. $\frac{8}{35}$
10. Randomly arrange the players in a row (this can be done in 8! ways) so that the two leftmost players compete against each other, the next two compete against each other, and so on. Now suppose we want no two mathematicians to play against each other. Then there are 8 possible positions for the first mathematician, then 6 positions for the second mathematician, and similarly 4 for the third and 2 for the last. After the positions of the mathematicians are fixed, there are 4! ways to arrange the positions of the non-mathematicians. Hence the answer is $\frac{8 \times 6 \times 4 \times 2 \times 4!}{8!}=\frac{8}{35}$.