Question

Eight people, including A and B, are to be seated around two identical tables, each having a capacity of $4$. The number of seating arrangement, so that A and B are not at the same table, is?

• A. $360$
• B. $1440$
• C. $720$
• D. $2880$

Answer 1

The correct option is C $720$

There are $8$ people including $A$ and $B$ wit $2$ table having each capacity of $4$.

As per question, $A$ and $B$ are on different tables.

So, by fixing $A$ on one table and $B$ on another table, we have remaining $6$ people divide into two groups of 3 each and this can be done in .

$\frac{6!}{3!3!2!}$ ways

But these groups can be assigned to $A$ and $B$ in two ways

Hence total possibility is $\frac{6!}{3!3!}$ ways

Now at each table there are $3$ people excluding $A$ and $B$ and they can be arranged in $3!$ ways

Hence total no of ways $=$\frac{6!}{3!3!}\times 3!\times 3!=720$