Number of ways of seating boys and girls around a table such that two particular boys always sit together and no two girls sit together

Question

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 ?

Answer 1

First seat the $4$ girls around the table, which can be done in $(4-1)!$ ways (why?, Check "circular permutation"). Now in between these girls, there are $4$ gaps, where you would have to seat the $5$ boys, two of whom must sit together. So we club the two boys to form a single unit and see that the $4$ units can be seated in the $4$ gaps in $4!$ ways. However the two boys who form the unit, can be rearranged among themselves in $2$ ways. So total number of ways is $2!\times3!\times4!$.

Answer 2

Considering 2 boys as a unit. (I will treat them as 4 boys)

Now there are 4 boys, around a circular table only 4 gaps will be made.(5 if it was a straight queue)

$4!$ Permutating 4 boys

$2!$ Internal permutation of 2 boys in the group

$3!$ Permutating 4 girls

$$3!2!4!$$