Question

Twelve persons are to be arranged around two round tables such that one table can accommodate seven persons and another table can accommodate five persons only.

The total number of possible arrangements if two particular persons $A$ and $B$ do not want to be on the same table is

• A. ${}^{10}{C}_4\times 6!\times 4!$
• B. $2\times {}^{10}{C}_6\times 6!\times 4!$
• C. ${}^{11}{C}_6\times 6!\times 4!$
• D. $2\times {}^{11}{C}_4\times 6!\times 4!$

Answer 1

The correct option is B $2\times {}^{10}{C}_6\times 6!\times 4!$

Here, $A$ can sit on first table and $B$ on the second
or
$A$ on second table and $B$ on the first table.

If $A$ is on the first table, then remaining $6$ for the first table can be selected in ${}^{10}{C}_6$ ways.
Now these $7$ persons can be arranged in $6!$ ways. Remaining $5$ can be arranged on the other table in $4!$ ways.

If $A$ is on the second table, then remaining $4$ for the second table can be selected in ${}^{10}{C}_4$ ways.
Now these $5$ persons can be arranged in $4!$ ways.
Remaining $7$ can be arranged on the other table in $6!$ ways.

Hence, total number of ways $=2\times {}^{10}{C}_6\times 6!\times 4!+2\times {}^{10}{C}_4\times 4!\times 6!=2\times {}^{10}{C}_6\times 6!\times 4!$