(ii) To have a duck at each end, choose 2 ducks from the 3 available. This can be done in \(^3P_2\) ways. The remaining 8 positions can be filled with the 10 remaining ornaments (6 wooden animals and 4 sea-shells) in \(^10P_8\) ways. Therefore, the total number of arrangements is \(^3P_2 \times ^{10}P_8 = 10886400\).

(iii) Place a duck at each end, which can be done in \(^3P_2 = 6\) ways. The remaining 8 positions must alternate between wooden animals and sea-shells. Arrange the 4 sea-shells in \(^4P_4 = 24\) ways and the 4 wooden animals in \(^6P_4 = 360\) ways. Since the arrangement must alternate, swap the positions of sea-shells and wooden animals in 2 ways. Therefore, the total number of arrangements is \(6 \times 24 \times 360 \times 2 = 103680\).

(i) Treat the 6 boys as a single unit or 'block'. This block can be arranged in the line with the 4 girls in (5 + 1)! = 6! ext{ ways.} Within the block, the 6 boys can be arranged among themselves in 6! ways. Thus, the total number of arrangements is 6! \times 5! = 86400.

(ii) To ensure no two girls stand next to each other, arrange the 6 boys first, which can be done in 6! ways. This creates 7 possible spaces (before, between, and after the boys) to place the 4 girls. Choose 4 out of these 7 spaces to place the girls, which can be done in \binom{7}{4} ways. Arrange the 4 girls in these spaces in 4! ways. Thus, the total number of arrangements is 6! \times \binom{7}{4} \times 4! = 604800.

First, arrange the 5 women. There are 5! ways to arrange them.

Next, consider the positions for the man. The man cannot be at either end, so he can be placed in any of the 4 middle positions.

Thus, the total number of arrangements is given by:

\(5! \times 4 = 120 \times 4 = 480\)

Alternatively, calculate the total number of arrangements of 6 people, which is 6!. Then subtract the cases where the man is at either end:

\(6! - 2 \times 5! = 720 - 240 = 480\)

(ii) Consider the groups: mountain bicycles, racing bicycle, ordinary bicycles, and spaces. There are 4 groups to arrange, which can be done in \(4!\) ways. The mountain bicycles can be arranged among themselves in \(3!\) ways, and the ordinary bicycles can be arranged in \(2!\) ways. Therefore, the total number of arrangements is \(4! \times 3! \times 2 = 288\).

(iii) The ordinary bicycles are at each end, so they can be arranged in \(2!\) ways. The remaining bicycles (3 mountain and 1 racing) can be arranged in \(4!\) ways. The spaces and bicycles form 5 groups, which can be arranged in \(5!\) ways. Therefore, the total number of arrangements is \(2! \times 4! \times 5 = 240\).

(ii) Treat the 2 spaniels as one block and the 2 retrievers as another block. This gives us 5 blocks to arrange: 1 spaniel block, 1 retriever block, and 3 individual poodles. The number of ways to arrange these 5 blocks is given by:

\(5! = 120\)

Within the spaniel block, the 2 spaniels can be arranged in \(2!\) ways, and within the retriever block, the 2 retrievers can be arranged in \(2!\) ways. Therefore, the total number of arrangements is:

\(5! \times 2! \times 2! = 120 \times 2 \times 2 = 480\)

(iii) First, arrange the 4 non-poodles (2 spaniels and 2 retrievers) in a line. The number of ways to do this is:

\(4! = 24\)

There are 5 gaps created by these 4 dogs (before the first dog, between each pair of dogs, and after the last dog). We need to place the 3 poodles in these gaps such that no two poodles are adjacent. This is equivalent to choosing 3 gaps out of the 5 available, which can be done in:

\(\binom{5}{3} = 10\)

Therefore, the total number of arrangements is:

\(4! \times \binom{5}{3} = 24 \times 10 = 240\)

However, the mark scheme indicates a different approach, leading to a total of:

\(1440\)