(i) There are 9 people in total. The number of possible arrangements with no restrictions is given by the factorial of the total number of people:

\(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\).

(ii) First, arrange the 6 men. The number of ways to do this is \(6!\):

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\).

There are 7 gaps created between and at the ends of the men where the women can be placed (before the first man, between men, and after the last man). We need to choose 3 of these 7 gaps to place the women, ensuring no two women are adjacent. The number of ways to choose 3 gaps from 7 is given by \(\binom{7}{3}\), and the number of ways to arrange the 3 women in these gaps is \(3!\):

\(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\).

\(3! = 3 \times 2 \times 1 = 6\).

The total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).