Question

$12$ persons are to be arranged around two round tables such that one table can accommodate seven persons and another five persons only. Then the number of ways of arrangement if two particular persons $A$ and $B$ want to be together and adjacent, is

• A. ${}^{10}{C}_{5}5!4!2!+{}^{10}{C}_{7}6!3!2!$
• B. ${}^{10}{C}_{5}5!4!+{}^{10}{C}_{7}6!3!$
• C. ${}^{10}{C}_{5}5!3!2!+{}^{10}{C}_{7}6!4!2!$
• D. ${}^{10}{C}_{5}5!4!2!+{}^{10}{C}_{7}5!3!2!$

Answer 1

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

Let first table and second table can accommodate seven persons and five persons respectively.
If $A$ and $B$ are on first table, then remaining five 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 $2!5!$ ways. Remaining $5$ can be arranged on the second table in $4!$ ways.
So, total number of ways is ${}^{10}{C}_{5}4!5!2!$.

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 $2!3!$ ways. Remaining $7$ can be arranged on the first table in $6!$ ways.
Hence, number of ways for first table is ${}^{10}{C}_{7}6!3!2!$.

Therefore, total number of ways is ${}^{10}{C}_{5}4!5!2!+{}^{10}{C}_{7}6!3!2!$.