(i) The word HAPPINESS consists of 9 letters where P appears twice and S appears twice. The number of different arrangements is given by:

\(\frac{9!}{2! \times 2!} = \frac{362880}{4} = 90720\)

(ii) To find the probability that the 3 vowels (A, E, I) are not all next to each other, first calculate the number of arrangements where they are together. Treat the 3 vowels as a single unit, so we have 7 units to arrange: (AEI), H, P, P, N, S, S.

The number of arrangements of these 7 units is:

\(\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260\)

Within the unit (AEI), the vowels can be arranged in \(3!\) ways:

\(3! = 6\)

Thus, the total number of arrangements where the vowels are together is:

\(1260 \times 6 = 7560\)

The probability that the vowels are not all together is:

\(1 - \frac{7560}{90720} = \frac{83160}{90720} = 0.917 \left(=\frac{11}{12}\right)\)

(i) To find the number of possible seating arrangements for the 12 passengers out of 14 seats, we use permutations:

\(^{14}P_{12} = \frac{14!}{(14-12)!} = \frac{14!}{2!} = 4.36 \times 10^{10}\).

(ii) For the specific seating arrangement:

1. The 3 business people can be arranged in the front row in \(3! = 6\) ways.
2. The 5 students each sit at a window seat, which can be arranged in \(5! = 120\) ways.
3. Mr and Mrs Lin can sit in the same row on the same side of the aisle in \(3P2 \times 2 = 12\) ways (choosing 2 out of 3 rows and 2 ways to arrange them).
4. Mr and Mrs Brown can sit in another row on the same side of the aisle in \(2 \times 2 = 4\) ways.
5. Total arrangements = \(6 \times 120 \times 12 \times 4 = 17280\).

(i) To find the number of different arrangements of 6 pegs, all of different colours, in 4 holes, we use permutations. The number of ways to choose 4 pegs from 6 is given by the permutation formula:

\(^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 360\)

Thus, there are 360 different arrangements.

(ii) For 4 pegs consisting of 2 blue pegs, 1 orange peg, and 1 yellow peg, we need to account for the repetition of the blue pegs. The formula for permutations of multiset is:

\(\frac{4!}{2!} = 12\)

Therefore, there are 12 different arrangements.

Consider Abel and Betty as a single unit or block. This reduces the problem to arranging 6 units: (AB), Cally, Doug, Eve, Freya, and Gino.

The number of ways to arrange these 6 units is given by:

\(6! = 720\)

Within the block (AB), Abel and Betty can be arranged in:

\(2! = 2\)

Thus, the total number of arrangements with Abel and Betty together is:

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

Now, consider the arrangements where Freya and Gino are also together. Treat Freya and Gino as another block, reducing the problem to arranging 5 units: (AB), (FG), Cally, Doug, and Eve.

The number of ways to arrange these 5 units is:

\(5! = 120\)

Within the block (FG), Freya and Gino can be arranged in:

\(2! = 2\)

Thus, the total number of arrangements with both (AB) and (FG) together is:

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

Therefore, the number of arrangements where Abel and Betty are together but Freya and Gino are not together is:

\(1440 - 480 = 960\)

(i) Consider the cases where the first and last cans are the same type:

• If they are cola, the arrangement is \\frac{5!}{2!2!} = 30\ ways.
• If they are green tea, the arrangement is \\frac{5!}{3!2!} = 10\ ways.
• If they are orange juice, the arrangement is \\frac{5!}{3!2!} = 10\ ways.

\(Total arrangements = 30 + 10 + 10 = 50\ ways.\)

\((ii) Treat the 3 cans of cola as a single block. The arrangement without restriction is \\frac{5!}{2!2!} = 30\ ways.\)

\(Now consider the case where both colas and green teas are together: \\frac{4!}{2!} = 12\ ways.\)

\(Subtract the restricted case from the total: 30 - 12 = 18\ ways.\)