Question

In how many ways $5$ girls and $4$ boys can be seated at a round table if all the $4$ boys do not sit together.

• A. $8!-(5!\times 4!)$
• B. $8!\times 4!$
• C. $9!$
• D. $5!\times 4!$

Answer 1

Answer: (A)

$8!-(5!\times 4!)$

Total number of arrangements $=(9-1)!=8!

Subtract the number of ways when $4$ boys when they are together

Take $4$ as single entity while they can also permuted among themselves.

$(6-1)!\times 4!$

$5!\times 4!$

Number of ways $=8!-$$5!\times 4!$