Number of ways in which 5 boys and 4 girls can be seated around a circular table such that no two girls sit together and two particular boys are always together ?
The answer to this question is $3!2!4!$ . It is done by considering $2$ boys as one unit and the the number of units (of boys) is $4$ so they can be arranged in $3!2!$ ways. Then number of girls can be arranged in $4!$ ways. However , I have a doubt. After five boys have been seated aren’t there $5$ places created for girls ? So girls should actually be seated in $\binom{5}{4}4!$ ways right ? So final answer should be $3!2!\binom{5}{4}4!$ right ?