(i) Freddie has a total of 9 toys (6 cars + 3 buses). The number of ways to choose 4 toys from 9 is given by the combination formula:

\(\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\)

(ii) If Freddie's choice must include both his favourite car and his favourite bus, he has already chosen 2 toys. He needs to choose 2 more toys from the remaining 7 toys (5 cars + 2 buses). The number of ways to choose 2 toys from 7 is:

\(\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21\)

We need to consider different scenarios based on the given conditions:

1. 4 Violinists (V), 1 Cellist (C), and 1 Double Bass player (DB):

   The number of ways to choose 4 violinists from 11 is given by \(\binom{11}{4}\).

   The number of ways to choose 1 cellist from 5 is given by \(\binom{5}{1}\).

   The number of ways to choose 1 double bass player from 4 is given by \(\binom{4}{1}\).

   Total ways for this scenario: \(\binom{11}{4} \times \binom{5}{1} \times \binom{4}{1} = 6600\).

2. 4 Violinists and 2 Cellists:

   The number of ways to choose 4 violinists from 11 is \(\binom{11}{4}\).

   The number of ways to choose 2 cellists from 5 is \(\binom{5}{2}\).

   Total ways for this scenario: \(\binom{11}{4} \times \binom{5}{2} = 3300\).

3. 5 Violinists and 1 Cellist:

   The number of ways to choose 5 violinists from 11 is \(\binom{11}{5}\).

   The number of ways to choose 1 cellist from 5 is \(\binom{5}{1}\).

   Total ways for this scenario: \(\binom{11}{5} \times \binom{5}{1} = 2310\).

Adding all scenarios together gives the total number of ways:

\(6600 + 3300 + 2310 = 12210\).

To divide 9 people into groups of 4, 3, and 2, we use combinations.

First, choose 4 people out of 9 for the first group: \(\binom{9}{4}\).

Then, choose 3 people out of the remaining 5 for the second group: \(\binom{5}{3}\).

The last 2 people automatically form the third group: \(\binom{2}{2}\).

Calculate each combination:

\(\binom{9}{4} = 126\)

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

\(\binom{2}{2} = 1\)

Multiply these results to find the total number of ways:

\(126 \times 10 \times 1 = 1260\)

To solve this problem, we need to calculate the number of ways to choose 4 pieces of jewellery such that at least 1 necklace and at least 1 bracelet are included.

We can use two methods to solve this:

Method 1:

1. Choose 1 necklace, 2 rings, and 1 bracelet: \(\binom{2}{1} \times \binom{8}{2} \times \binom{4}{1} = 2 \times 28 \times 4 = 224\)
2. Choose 1 necklace, 1 ring, and 2 bracelets: \(\binom{2}{1} \times \binom{8}{1} \times \binom{4}{2} = 2 \times 8 \times 6 = 96\)
3. Choose 2 necklaces, 1 ring, and 1 bracelet: \(\binom{2}{2} \times \binom{8}{1} \times \binom{4}{1} = 1 \times 8 \times 4 = 32\)
4. Choose 2 necklaces and 2 bracelets: \(\binom{2}{2} \times \binom{4}{2} = 1 \times 6 = 6\)
5. Choose 1 necklace and 3 bracelets: \(\binom{2}{1} \times \binom{4}{3} = 2 \times 4 = 8\)

\(Total = 224 + 96 + 32 + 6 + 8 = 366 ways.\)

Method 2:

Calculate the total number of ways to choose 4 pieces from 14 items (2 necklaces + 8 rings + 4 bracelets) and subtract the cases where there are no necklaces or no bracelets:

Total ways: \(\binom{14}{4} = 1001\)

Subtract cases with no necklaces (only rings and bracelets): \(\binom{12}{4} = 495\)

Subtract cases with no bracelets (only necklaces and rings): \(\binom{10}{4} = 210\)

Subtract cases with only rings: \(\binom{8}{4} = 70\)

Using inclusion-exclusion principle, we find:

\(1001 - (495 + 210 - 70) = 366\)

Thus, the number of possible selections is 366.

To find the number of selections that include cars of at least 2 different colours, we consider all possible combinations of selecting 3 cars from the 12 available, and then subtract the combinations where all selected cars are of the same colour.

First, calculate the total number of ways to select 3 cars from 12:

\(\binom{12}{3} = 220\)

Next, calculate the number of ways to select 3 cars of the same colour:

• All red: \(\binom{5}{3} = 10\)
• All white: \(\binom{4}{3} = 4\)
• All black: \(\binom{3}{3} = 1\)

Total ways to select 3 cars of the same colour:

\(10 + 4 + 1 = 15\)

Therefore, the number of ways to select 3 cars with at least 2 different colours is:

\(220 - 15 = 205\)