Question

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

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

Answer 1

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

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