(ii) Treat each group of trees (hibiscus, jacaranda, oleander) as a single unit or 'super tree'. This gives us 3 'super trees' to arrange. The number of ways to arrange these 3 'super trees' is given by:

\(3!\)

Within each 'super tree', the trees can be arranged internally. The number of ways to arrange the hibiscus trees is \(4!\), the jacaranda trees is \(6!\), and the oleander trees is \(2!\). Therefore, the total number of arrangements is:

\(3! \times 4! \times 6! \times 2! = 207360\)

(iii) First, arrange the 8 non-hibiscus trees (6 jacaranda and 2 oleander) in a line. The number of ways to do this is:

\(8!\)

This creates 9 gaps (including the ends) where the hibiscus trees can be placed. We need to choose 4 out of these 9 gaps to place the hibiscus trees, ensuring no two hibiscus trees are adjacent. The number of ways to choose 4 gaps from 9 is given by:

\(\binom{9}{4}\)

Finally, arrange the 4 hibiscus trees within the chosen gaps. The number of ways to do this is:

\(4!\)

Therefore, the total number of arrangements is:

\(8! \times \binom{9}{4} \times 4! = 121,927,680\)

(i) Consider each family as a single unit or 'block'. There are 4 families, so we arrange these 4 blocks in different orders:

\(4! = 24 ext{ ways}\)

Within each family, the members can be arranged among themselves. The number of arrangements for each family is:

2! ext{ for the Lin family,}

3! ext{ for the O’Connor family,}

5! ext{ for the Ahmed family,}

2! ext{ for the Burton family.}

Thus, the total number of arrangements is:

\(4! imes 2! imes 3! imes 5! imes 2! = 829440\)

(ii) First, arrange the 8 children. There are:

8! ext{ ways to arrange the children.}

To ensure no two adults are consecutive, place the 6 adults in the 9 gaps created by the 8 children (before the first child, between children, and after the last child). Choose 6 out of these 9 gaps:

9 imes 8 imes 7 imes 6 imes 5 imes 4 ext{ ways to choose and arrange the gaps.}

Thus, the total number of arrangements is:

\(8! imes 9 imes 8 imes 7 imes 6 imes 5 imes 4 = 2438553600\)

(b) Treat each group (swimmers, cyclists, runners) as a single unit. This gives us 3 units to arrange, which can be done in \(3!\) ways. Within each group, the swimmers can be arranged in \(2!\) ways, the cyclists in \(3!\) ways, and the runners in \(4!\) ways. Therefore, the total number of arrangements is \(3! \times 2! \times 3! \times 4! = 6 \times 2 \times 6 \times 24 = 1728\).

(c) First, arrange the 6 non-cyclists (2 swimmers and 4 runners) in \(6!\) ways. This creates 7 gaps (before, between, and after these 6 people) where the 3 cyclists can be placed. Choose 3 out of these 7 gaps to place the cyclists, which can be done in \(\binom{7}{3}\) ways. The cyclists themselves can be arranged in \(3!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(i) The number of ways to arrange 8 people in 12 seats with no restrictions is given by the permutation formula:

\(^{12}P_8 = \frac{12!}{(12-8)!} = 19,958,400\).

(ii) To ensure Mary and Frances do not sit next to each other, first calculate the total arrangements if they sit together. Treat Mary and Frances as a single unit, so there are 7 units to arrange in 11 seats:

\(^{11}P_7 = \frac{11!}{(11-7)!} = 1,663,200\).

Since Mary and Frances can switch places, multiply by 2:

\(1,663,200 \times 2 = 3,326,400\).

Subtract this from the total unrestricted arrangements:

\(19,958,400 - 3,326,400 = 16,632,000\).

(iii) If all 8 people sit together, treat them as a single block. There are 5 possible positions for this block in 12 seats. The number of ways to arrange the 8 people within the block is:

\(8! = 40,320\).

Multiply by the 5 positions:

\(40,320 \times 5 = 201,600\).

(a) Consider the packets of chocolates as a single unit. This gives us 5 packets of toffees, 4 packets of fruit gums, and 1 unit of chocolates, totaling 10 units to arrange. The number of arrangements is given by:

\(\frac{10!}{5!4!} = 1260\)

(b)(i) Jessica chooses 4 packets out of 8, considering the order. This is a permutation problem:

\({}_8P_4 = 8 \times 7 \times 6 \times 5 = 1680\)

(b)(ii) If the packets of chocolate biscuits and custard creams are chosen, we need to choose 2 more packets from the remaining 6. The number of ways to choose 2 packets from 6 is:

\(\binom{6}{2} = 15\)

Considering the order of 4 packets, we multiply by \(4!\):

\(15 \times 4! = 360\)