How Are Permutations Calculated in a Circular Seating Arrangement?

[Happiness]

2 boys and 3 girls are to be seated round a table with 5 seats. Each child occupies exactly one seat. In how many ways can this be done if
(a) the 2 boys must be seated together
(b) same as (a) but this time the seats are numbered

Solution
(a) $\frac{4!}{4}2!$
(b) $\frac{4!}{4}2!\times 5$

The "$\times 5$" in (b) seems to contradict the "$\div 4$" in (a).

Why ain't the answers
(a) $4!2!\div 5$, (b) $4!2!$ or
(a) $\frac{4!}{4}2!$, (b) $4!2!$

Explanation for the "correct" solution
(a) After grouping the 2 boys into one group, we have 4 groups (with one girl in a group). The number of ways of arranging 4 items in a circle is $\frac{4!}{4}$ since ABCD, DABC, CDAB and BCDA are the same arrangement when placed in a circle. The 2 boys can be permutated within the "boy group". So we have $\frac{4!}{4}2!$.
(b) If the seats are numbered, rotating an arrangement by one seat results in a new arrangement. We can do 5 such rotations. So we have $\frac{4!}{4}2!\times 5$.

But if we $\div 4$ in (a) since we are rotating the 4 groups ABCD 4 times and considering these 4 arrangements to be the same, then in (b) shouldn't we $\times 4$ instead of $5$?

On the other hand, if we first consider the seats to be numbered and hence ABCD, DABC, CDAB and BCDA are NOT the same arrangement, we have 4!2!. Then for (a), since the seats are not numbered and there are 5 rotations that produce the same arrangement when there are 5 seats, we have (a) $4!2!\div 5$.

My issue is why does the correct answer consider the rotations of groups in (a) but not in (b)? In (b), it considers instead the rotations of seats (or persons).

 

[mathman]

The grouping description seems to unduly complicate the problem. 2 boys have 2! possible permutations, 3 girls have 3! possible permutations, so the seating when the seats are not numbered is 2!3! - the 2 boys together and the 3 girls together. Seats being numbered multiplies by 5, since there are 5 possible positions for girl in the middle.

 

[Happiness]

The grouping description seems to unduly complicate the problem. 2 boys have 2! possible permutations, 3 girls have 3! possible permutations, so the seating when the seats are not numbered is 2!3! - the 2 boys together and the 3 girls together. Seats being numbered multiplies by 5, since there are 5 possible positions for girl in the middle.

But this method can't be generalised while the grouping method can be.

2 boys, 3 girls and 4 men are to be seated round a table with 9 seats. In how many ways can this be done if
(a) the 2 boys must be seated together and the 4 men must be seated together
(b) same as (a) but this time the seats are numbered

Solution
(a) Grouping the boys into one group and the men into another, we have 5 groups. So the answer is $\frac{5!}{5}2!4!$.
(b) Considering the 9 rotations of seats, we have $\frac{5!}{5}2!4!\times 9$.

We have the same issue: we consider rotations of groups when we $\div 5$ in (a) but we consider rotations of seats when we $\times 9$ in (b).