Permutations Seating Arrangement in Group

Question

As per the question, I think the answer is $2\times4! = 48$ ways, but unable to visualize how they are seated in?

Answer 1

There are six seats: $1-2-3-4-5-6$.

Let's call Mr. Spencer $S$, his two children are $C_1$ and $C_2$.

There are $8$ cases:

$$C_1-S-C_2-...-...-...$$

$$C_2-S-C_1-...-...-...$$

$$...-C_1-S-C_2-...-...$$

$$...-C_2-S-C_1-...-...$$

$$...-...-C_1-S-C_2-...$$

$$...-...-C_2-S-C_1-...$$

$$...-...-...-C_1-S-C_2$$

$$...-...-...-C_2-S-C_1$$

For each case, the other $3$ people sit randomly (anywhere is fine), so there are $3!=6$ possible ways for each case.

There are $8\times 3!=48$ ways overall, so your guess is correct.