Question

In a family there are $10$ adults and $6$ children. In how many ways can they be seated around table so that no two of the children sit together?

• A. $15!$
• B. $9!\times \frac{10!}{4!}$
• C. $10!\times \frac{11!}{5!}$
• D. $16!$

Answer 1

The correct option is B $9!\times \frac{10!}{4!}$

$10$ adults

$6$ children

Concider a round table with $10$ places

Let the adult be selected first

Number of ways they can be selected

$\Rightarrow (10-1)!\Rightarrow 9!$

Now let the children be placed in between them

There are 10 possible places and 6 children

Number of ways of aranging

${\Rightarrow }^{10}{P}_6\Rightarrow \frac{10!}{4!}$

Total number of ways of arranging

$\Rightarrow 9!\times \frac{10!}{4!}$

Option $:B$