(i) The total number of arrangements of 9 cards is given by the factorial of 9, which is:

\(9! = 362880\)

(ii) To find the number of arrangements where the pink and green cards are not next to each other, first calculate the total arrangements where they are together. Treat the pink and green cards as a single unit, so there are 8 units to arrange:

\(8! = 40320\)

Within this unit, the pink and green cards can be arranged in 2 ways (PG or GP):

\(8! \times 2 = 80640\)

Subtract this from the total arrangements:

\(362880 - 80640 = 282240\)

(iii) The number of ways to choose and arrange 3 cards from 9 is given by:

\(^9P_3 = \frac{9!}{(9-3)!} = 504\)

(iv) To find the number of arrangements containing the pink card, choose 2 more cards from the remaining 8 and arrange them:

\(^8C_2 \times 3! = 28 \times 6 = 168\)

(v) For arrangements not having the pink card next to the green card, calculate the arrangements where they are together and subtract from the total in part (iii). Treat PG as a single unit, choose 1 more card from the remaining 7, and arrange:

\(7 \times 2 \times 2 = 28\)

Subtract from total arrangements of 3 cards:

\(504 - 28 = 476\)

(ii) To arrange the 9 mugs such that the 3 china mugs are separated, first arrange the 6 plastic mugs. There are \(6!\) ways to arrange these. This creates 7 gaps (before, between, and after the plastic mugs) to place the china mugs. Choose 3 out of these 7 gaps, which can be done in \(\binom{7}{3}\) ways. Arrange the 3 china mugs in these gaps in \(3!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(iii) Treat the 3 identical red mugs as a single unit. This gives us 12 units to arrange: 1 red unit, 4 blue mugs, and 7 yellow mugs. The total number of arrangements is \(\frac{12!}{4! \times 7!}\), accounting for the identical blue and yellow mugs. Therefore, the number of arrangements is \(\frac{479001600}{24 \times 5040} = 3960\).

(ii) Consider the 6 tallest sopranos as a single block. This block can be arranged in a line with the remaining 4 sopranos. There are 5 blocks in total (1 block of 6 sopranos and 4 individual sopranos). The number of ways to arrange these 5 blocks is given by:

\(5!\)

Within the block of 6 sopranos, they can be arranged among themselves in:

\(6!\)

Therefore, the total number of arrangements is:

\(5! \times 6! = 86400\)

(iii) Consider the 4 tenors as a single block and the 4 basses as another block. These two blocks can be arranged in:

\(2!\)

Within the block of 4 tenors, they can be arranged among themselves in:

\(4!\)

Within the block of 4 basses, they can be arranged among themselves in:

\(3!\)

Therefore, the total number of arrangements is:

\(4! \times 3! \times 2 = 288\)

(i) To find the number of possible arrangements of the 12 houses, we use the formula for permutations of multiset:

\(\frac{12!}{2!2!3!4!}\)

Calculating this gives:

\(\frac{479001600}{4 \times 2 \times 6 \times 24} = 831600\)

(ii) For the second part, we need to arrange the houses in two groups of 6. In the first group, we have all houses in styles A and D, which are 2 and 4 respectively. The number of arrangements for the first group is:

\(\frac{6!}{4!2!}\)

In the second group, we have all houses in styles B, C, and E, which are 2, 3, and 1 respectively. The number of arrangements for the second group is:

\(\frac{6!}{2!3!}\)

The total number of arrangements is the product of the arrangements of the two groups:

\(\frac{6!}{4!2!} \times \frac{6!}{2!3!} = 900\)

To solve this problem, treat the 3 jazz CDs as a single unit or 'block'. This reduces the problem to arranging 9 units: 6 pop CDs, 2 classical CDs, and 1 jazz block.

The number of ways to arrange these 9 units is given by:

\(9!\)

Within the jazz block, the 3 jazz CDs can be arranged among themselves in:

\(3!\)

Therefore, the total number of arrangements is:

\(9! \times 3!\)

Calculating these factorials:

\(9! = 362,880\)

\(3! = 6\)

Thus, the total number of arrangements is:

\(362,880 \times 6 = 2,177,280\)

Alternatively, the mark scheme suggests another possible answer of 2,180,000, which might be due to rounding or approximation in a different context.