(i) Consider Jai and Kaz as a single unit or block. Similarly, each friend and his wife can be considered as a block. This results in 6 blocks to arrange. The number of ways to arrange these 6 blocks is given by:

\(6!\)

Within each block, the two people can be arranged in \(2!\) ways. Therefore, the total number of arrangements is:

\(6! \times 2^6 = 720 \times 64 = 46080\)

(ii) Jai and Kaz occupy the two middle positions, so they are fixed. The five friends can be arranged on one side in \(5!\) ways, and the five wives can be arranged on the other side in \(5!\) ways. Therefore, the total number of arrangements is:

\(5! \times 5! \times 2 \times 2 = 120 \times 120 \times 4 = 57600\)

(a) With a red candle at each end, we have the arrangement R _ _ _ _ _ _ _ _ _ R. This leaves 9 positions to fill with 3 blue and 6 green candles. The number of arrangements is given by:

\(\frac{9!}{3!6!} = 84\)

(b) Treat the 3 blue candles as a single block, so we have B B B as one unit. This leaves us with 7 units to arrange: B B B, G, G, G, G, G, G, and 2 red candles. First, arrange the 7 units:

\(\frac{7!}{6!} = 7\)

Now, consider the arrangements where the red candles are not together. Calculate the total arrangements of the 7 units and subtract the arrangements where the red candles are together:

Total arrangements with blues together: \(\frac{9!}{6!} = 84\)

Arrangements with blues and reds together: \(\frac{8!}{6!} = 56\)

So, the number of arrangements where the red candles are not together is:

\(252 - 56 = 196\)

(i) Consider Mr Keene and Mr Uzuma fixed at the two ends of the line. This leaves 9 people to arrange in between them. The number of ways to arrange 9 people is given by the factorial of 9, which is:

\(9! = 362880\)

Since Mr Keene and Mr Uzuma can switch places, we multiply by 2:

\(9! \times 2 = 725760\)

Thus, the number of ways is 725760.

(ii) Treat the 5 Keene children as a single unit and the 2 Uzuma children as another single unit. This gives us 7 units to arrange: 5 Keene children, 2 Uzuma children, Mr and Mrs Keene, and Mr and Mrs Uzuma. The number of ways to arrange these 7 units is:

\(7! = 5040\)

Within the Keene children unit, the 5 children can be arranged among themselves in:

\(5! = 120\)

Within the Uzuma children unit, the 2 children can be arranged among themselves in:

\(2! = 2\)

Thus, the total number of arrangements is:

\(7! \times 5! \times 2! = 5040 \times 120 \times 2 = 172800\)

Therefore, the number of ways is 172800.

(iii) Treat the 3 buses as a single unit. This gives us 7 units to arrange: 6 cars and 1 bus unit. The number of ways to arrange these 7 units is given by:

\(7!\)

Within the bus unit, the 3 buses can be arranged among themselves in:

\(3!\)

Thus, the total number of arrangements is:

\(7! \times 3! = 5040 \times 6 = 30240\)

(iv) Place a car at each end, leaving 4 cars and 3 buses to arrange in the middle. The number of ways to arrange 4 cars and 3 buses such that no two buses are adjacent is equivalent to arranging the 4 cars first and then placing the buses in the gaps between and around the cars. There are 5 gaps (before, between, and after the cars) to place the buses:

Choose 3 out of these 5 gaps to place the buses:

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

Arrange the 4 cars:

\(4!\)

Arrange the 3 buses:

\(3!\)

Thus, the total number of arrangements is:

\(4! \times \binom{5}{3} \times 3! = 24 \times 10 \times 6 = 1440\)

Multiply by 6 for the 6 possible cars that can be at each end:

\(6 \times 1440 = 8640\)

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

\(6! \times 5P3 = 43200\)

To solve this problem, we can use two methods:

Method 1:

First, arrange the 5 men. There are 5! ways to do this.

\(5! = 120\)

Now, consider the gaps between the men where the women can be placed. There are 6 gaps (before the first man, between the men, and after the last man).

We need to choose 2 out of these 6 gaps to place the women, which can be done in \(^6P_2\) ways.

\(^6P_2 = 6 \times 5 = 30\)

Thus, the total number of ways is:

\(5! \times ^6P_2 = 120 \times 30 = 3600\)

Method 2:

Calculate the total number of arrangements without any restrictions, which is 7! (since there are 7 people).

\(7! = 5040\)

Calculate the number of arrangements where the women are together. Treat the two women as a single unit, so we have 6 units to arrange (5 men + 1 unit of women).

There are 6! ways to arrange these units, and within the unit of women, the 2 women can be arranged in 2! ways.

\(6! \times 2! = 720 \times 2 = 1440\)

The number of arrangements where the women are not together is:

\(7! - (6! \times 2!) = 5040 - 1440 = 3600\)

Both methods give the same result: 3600 ways.