Question

Twelve persons are to be arranged around two round tables such that one table can accommodate seven persons and another five persons only. The number of ways of arrangements if two particular persons $A$ and $B$ do not want to be on the same table is

• A. ${}^{10}{C}_{4}6!4!$
• B. $2{}^{10}{C}_{6}6!4!$.
• C. ${}^{11}{C}_{6}6!4!$.
• D. none of these

Answer 1

The correct option is B $2{}^{10}{C}_{6}6!4!$.

Here, $A$ can sit on first table and $B$ on the second or $A$ on the second table and $B$ on the first table.
If $A$ is on the first table, then remaining six for first table can be selected in ${}^{10}{C}_6$ ways. Now thses seven persons can be arranged in $6!$ ways. Remaining five can be arranged on the other table in $4!$ ways.
Hence, total number of ways is $2{}^{10}{C}_{6}6!4!$.