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 ways in which these $12$ persons can be arranged is

• A. ${}^{12}{C}_7\times 6!\times 4!$
• B. $6!\times 4!$
• C. ${}^{13}{C}_5\times 6!\times 4!$
• D. $2{}^{12}{C}_7\times 6!\times 4!$

Answer 1

The correct option is A ${}^{12}{C}_7\times 6!\times 4!$

$7$ persons can be selected for first table in${}^{12}{C}_7$ ways.

Now these $7$ persons can be arranged in $6!$ ways.

The remaining $5$ persons can be arranged on the second table in $4!$ ways.

Hence, total number of ways
$={}^{12}{C}_7\times 6!\times 4!$