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 arrangement if two particular persons A and B want to be together and consecutive is

• A. ${}^{10}{C}_{7}6!3!2!+{}^{10}{C}_{5}4!5!2!$
• B. ${}^{10}{C}_{5}6!3!+{}^{10}{C}_{7}4!5!$
• C. ${}^{10}{C}_{7}6!2!+{}^{10}{C}_{5}5!2!$
• D. none of these

Answer 1

The correct option is A ${}^{10}{C}_{7}6!3!2!+{}^{10}{C}_{5}4!5!2!$

If $A,B$ are on the first table, then remaining five can be selected in ${}^{10}{C}_5$ ways.
Now seven persons including $A$ and $B$ can be arranged on the first table in which $A$ and $B$ are together in $2!5!$ ways. Remaining five can be arranged on the second table in $4!$ ways. Total number of ways is ${}^{10}{C}_{5}4!5!2!$.
If $A,B$ are on the second table, then remaining three can be selected in ${}^{10}{C}_3$ ways.
Now five persons including $A$ and $B$ can be arranged on the second table in which $A$ and $B$ are together in $2!3!$ ways.
Remaining seven can be arranged on the first table in $6!$ ways.
Hence, number of ways for second case is ${}^{10}{C}_{7}6!3!2!.$