Question

$8$ boys are supposed to be seated around a circular table. The number of ways of arranging them around the table such that two particular boys should always seat together and other two particular boys should never seat together is

• A. $4560$
• B. $960$
• C. $1200$
• D. $3600$

Answer 1

The correct option is B $960$

Let the two boys who want to seat together be $A$ and $B$ and the other two be $X$ and $Y.$
Consider $A$ and $B$ as one entity so that they always seat together.
So number of ways of arranging them around a circular table is
$(7-1)!\times 2!$
Above condition is the case where $A$ and $B$ are always together but $X$ and $Y$ shouldn't be together.
So subtracting the case of $X$ and $Y$ always together from the above case will give the case where $A$ and $B$ are always together and $X$ and $Y$ are never together.
So required no. of ways $=(7-1)!\times 2!-(6-1)!\times 2!\times 2!$
$=960$