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 $B$ want to be together is

• A. ${}^{10}{C}_7\times 6!\times 3!\times 2!+{}^{10}{C}_5\times 4!\times 5!\times 2!$
• B. ${}^{10}{C}_5\times 6!\times 3!+{}^{10}{C}_7\times 4!\times 5!$
• C. ${}^{10}{C}_7\times 6!\times 2!+{}^{10}{C}_5\times 5!\times 2!$
• D. ${}^{10}{C}_5\times 6!\times 3!\times 2!+{}^{10}{C}_7\times 4!\times 5!$

Answer 1

The correct option is A ${}^{10}{C}_7\times 6!\times 3!\times 2!+{}^{10}{C}_5\times 4!\times 5!\times 2!$

$\left(i\right)$ If $A,B$ are on the first table, then
remaining $5$ can be selected in${}^{10}{C}_5$ ways.

Now $7$ persons including $A$ and $B$ can be arranged on the first table in
which $A$ and $B$ are together in $2!\times 5!$
ways.
Remaining $5$ can be arranged on the second table in $4!$ ways.
Total number of ways is $={}^{10}{C}_5\times 4!\times 5!\times 2!$

$\left(ii\right)$ If $A,B$ are on the second table, then remaining $3$ can be selected in ${}^{10}{C}_3$ ways.

Now $5$ persons including $A$ and $B$ can be arranged on the second table in which $A$ and $B$ are together in $2!\times 3!$ ways.
Remaining $7$ can be arranged on the first table in $6!$ ways.
Hence, total number of ways$={}^{10}{C}_7\times 6!\times 3!\times 2!$
So the total answer is$={}^{10}{C}_5\times 4!\times 5!\times 2!+{}^{10}{C}_7\times 6!\times 3!\times 2!$