(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 \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(i) To find the number of possible seating arrangements for the 11 passengers, we use the permutation formula \(^nP_r\), where \(n\) is the total number of seats and \(r\) is the number of passengers. Here, \(n = 17\) and \(r = 11\).

\(^{17}P_{11} = \frac{17!}{(17-11)!} = \frac{17!}{6!}\)

Calculating this gives \(4.94 \times 10^{11}\).

(ii) If 5 particular people sit in the back row, we first arrange these 5 people in the 5 seats of the back row, which can be done in \(5!\) ways.

Then, we arrange the remaining 6 passengers in the remaining 12 seats. This can be done in \(^{12}P_6\) ways.

Thus, the total number of arrangements is \(5! \times ^{12}P_6\).

\(5! = 120\)

\(^{12}P_6 = \frac{12!}{(12-6)!} = \frac{12!}{6!}\)

Calculating this gives \(79800000\).

(i) To find the number of different arrangements for parking the 9 cars and leaving 4 empty spaces, we use permutations. The total number of ways to arrange 9 cars in 13 spaces is given by the permutation formula:

\(^{13}P_9 = \frac{13!}{(13-9)!} = \frac{13!}{4!} = 259,459,200\).

(ii) If the 4 empty spaces are next to each other, treat them as a single block. This reduces the problem to arranging 10 blocks (9 cars + 1 block of empty spaces). The number of arrangements is:

\(10! = 3,628,800\).

(iii) The probability that there will not be 4 empty spaces next to each other is the complement of the probability that they are next to each other. Thus,

\(P(\text{not next to each other}) = 1 - \frac{\text{arrangements with 4 empty spaces together}}{\text{total arrangements}}\).

\(P = 1 - \frac{3,628,800}{259,459,200} = 0.986\).

Since Mr Lizo goes first, we consider the remaining families as groups:

• Lizo family: 4 members (excluding Mr Lizo)
• Kenny family: 6 members
• Martin family: 2 members
• Nantes family: 2 members

First, calculate the number of ways to arrange the families:

\(3! = 6 ext{ ways (for Kenny, Martin, Nantes)}\)

Next, calculate the arrangements within each family:

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

\(6! = 720 ext{ ways (Kenny family)}\)

\(2! = 2 ext{ ways (Martin family)}\)

\(2! = 2 ext{ ways (Nantes family)}\)

Multiply these together to find the total number of arrangements:

\(4! imes 6! imes 2! imes 2! imes 3! = 24 imes 720 imes 2 imes 2 imes 6 = 414,720\)

Consider the 3 boys as a single unit or 'block'. This reduces the problem to arranging 10 units: 1 block of boys, 4 girls, and 5 adults.

Since an adult must be at each end, we choose 2 adults to be at the ends. The number of ways to choose 2 adults from 5 is given by:

\(^5P_2 = 5 \times 4 = 20\)

Now, arrange the remaining 8 units (1 block of boys, 4 girls, and 3 adults) in the middle:

\(8! = 40,320\)

Within the block of boys, the 3 boys can be arranged among themselves in:

\(3! = 6\)

Therefore, the total number of arrangements is:

\(8! \times 3! \times ^5P_2 = 40,320 \times 6 \times 20 = 4,838,400\)