8 teachers and 4 students are sitting around a circular table at random, then the probability that no two…
8 teachers and 4 students are sitting around a circular table at random, then the probability that no two students sit together is:
$\frac{7}{88}$
$\frac{14}{33}$
$\frac{8}{33}$
$\frac{7}{33}$
Solution
Total number of arrangements of 8 teachers and 4 students in circular table $=(8+4-1)!=11$ !
Number of arrangements of 8 teachers and 4 students in circular table such that no two students sit together $={ }^8 C_4 \cdot 4!(8-1)!={ }^8 C_4 \cdot 4!\cdot 7!$
$\therefore$ Required probability
$=\frac{{ }^8 C_4 \cdot 4!\cdot 7!}{11!}=\frac{\frac{8 \cdot 7 \cdot 6 \cdot 5}{4!} \cdot 4!\cdot 7!}{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}=\frac{7}{33} .$