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 do not sit together?

• A. $16!$
• B. $15!$
• C. $16!-(11!\times 6!)$
• D. $15!-(10!\times 6!)$

Answer 1

The correct option is D $15!-(10!\times 6!)$

Total number of arrangements $=15!$

For finding no of ways , when all children do not sit together

$=15!-$ No of ways when all children sit together

When children will be together , $6$ children is considered as a entity and also these $6$ can be arranged among themselves.

Number of ways $=10!\times 6!$

For finding no of ways, when all children do not sit together

$=15!-$ $10!\times 6!$