Question

In a family there are $10$ adults and $6$ children. In how many ways can they be seated around table so that all the children sit together?

• A. $10!\times 6!$
• B. $11!\times 6!$
• C. $10{!}^2$
• D. $9!\times 6!$

Answer 1

The correct option is A $10!\times 6!$

Number of adults $=10$

Number of children$=6$

Circular arrangement

Consider children as one unit and each adult as one unit

Number of initial numbers of children $=6!$

Number of arrangements of $11$ units $=(11-1)!=10!$

Total possible arrangements

$\Rightarrow 10!\times 6!$

Option $:A$