A permutation problem involving round a table

Question

The question:

There are $4$ man, $2$ woman and $1$ child sitting in a round table. Find the number of ways of arranging the $7$ people if the child is seated:

i) Between two women.

ii) Between two men.

My attempt:

I've drawn a diagram to help me visualize it first:

I've assumed that there are $7$ seats only so therefore $7! = 5040$.

For
i) $\frac{7!}{4!} = 210$ ways.
ii) $\frac{7!}{3!} = 840$ ways.

However I'm not confident in my answer, I feel like something is missing. Can someone point out what I've done wrong here?

Answer 1

i) between two women

You're placing a woman on either side of 1 child, and since we have 2 women, there are 2 ways for this to occur: 2 x 1 = 2 (note: the child will always be in the middle)

There are 4 seats left for 4 men to sit with no conditions, so 4! = 24

Hence: there are 2 x 24 = 48 arrangements

ii) between two men

2 out of these 4 men are to be seated around the child, hence 4C2 = 6

Since there are 2 ways each of these 2 chosen men can be ordered, we calculate 6 x 2 = 12

We are left with 2 men, and 2 women who can be arranged with no conditions, hence 4! = 24

Altogether we have: 24 x 12 = 288 arrangements

Hope that helps :)