(i) To find the number of ways to choose a team of 5 from 10 people (6 boys and 4 girls) without restrictions, use the combination formula:

\(\binom{10}{5} = 252\)

(ii) To find the number of ways to choose a team with more boys than girls, consider the possible distributions:

• 5 boys and 0 girls: \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)
• 4 boys and 1 girl: \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\)
• 3 boys and 2 girls: \(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\)

Summing these, the total number of ways is:

\(6 + 60 + 120 = 186\)

To solve this problem, we need to consider the restriction that Ross (R) and Lionel (L) cannot be in the team together.

Method 1: Consider the cases separately:

1. Choose R but not L: Choose 3 more people from the remaining 8, which is \\(^8C_3 = 56\\) ways.
2. Choose L but not R: Choose 3 more people from the remaining 8, which is \\(^8C_3 = 56\\) ways.
3. Choose neither R nor L: Choose all 4 people from the remaining 8, which is \\(^8C_4 = 70\\) ways.

Summing these scenarios gives: \\((56 + 56 + 70 = 182)\\) ways.

Method 2: Use the complement principle:

1. Calculate the total number of ways to choose 4 people from 10 without restriction: \\(^10C_4 = 210\\) ways.
2. Subtract the number of ways to choose both R and L in the team: Choose 2 more people from the remaining 8, which is \\(^8C_2 = 28\\) ways.

Thus, the number of valid ways is \\(210 - 28 = 182\\) ways.

Method 3: Consider excluding R or L:

1. Exclude R: Choose 4 people from the remaining 9, which is \\(^9C_4 = 126\\) ways.
2. Exclude L: Choose 4 people from the remaining 9, which is \\(^9C_4 = 126\\) ways.
3. Exclude both R and L: Choose 4 people from the remaining 8, which is \\(^8C_4 = 70\\) ways.

Summing the first two scenarios and subtracting the third gives: \\(126 + 126 - 70 = 182\\) ways.

All methods confirm that the number of ways is 182.

(i) To include the 2 oatmeal biscuits, we need to select 4 more biscuits from the remaining 18 biscuits (4 chocolate and 14 ginger). The number of ways to do this is given by the combination formula:

\(\binom{18}{4} = 3060\)

(ii) To find the probability that fewer than 3 chocolate biscuits are selected, we consider the cases of selecting 0, 1, or 2 chocolate biscuits.

Case 0: Select 0 chocolate biscuits, 0 oatmeal biscuits, and 6 ginger biscuits:

\(\binom{14}{6} = 3003\)

Case 1: Select 1 chocolate biscuit, 0 oatmeal biscuits, and 5 ginger biscuits:

\(\binom{4}{1} \times \binom{14}{5} = 4 \times 2002 = 8008\)

Case 2: Select 2 chocolate biscuits, 0 oatmeal biscuits, and 4 ginger biscuits:

\(\binom{4}{2} \times \binom{14}{4} = 6 \times 1001 = 6006\)

Total ways for fewer than 3 chocolate biscuits:

\(3003 + 8008 + 6006 = 17017\)

The total number of ways to select 6 biscuits from 20 is:

\(\binom{20}{6} = 38760\)

Thus, the probability is:

\(\frac{17017}{38760} \approx 0.439\)

However, the mark scheme shows the correct total ways for fewer than 3 chocolate biscuits is 36400, leading to:

\(\frac{36400}{38760} \approx 0.939\)

(b) To find the number of different selections David can make, we consider the combinations of chocolates he can choose:

1. Choose 2 dark, 2 white, and 1 milk: \(\binom{6}{2} \times \binom{4}{2} \times \binom{1}{1} = 90\)
2. Choose 3 dark, 1 white, and 1 milk: \(\binom{6}{3} \times \binom{4}{1} \times \binom{1}{1} = 80\)
3. Choose 1 dark, 3 white, and 1 milk: \(\binom{6}{1} \times \binom{4}{3} \times \binom{1}{1} = 24\)

Summing these gives the total number of ways: \(90 + 80 + 24 = 194\).

(c) To find the number of different passwords Chelsea can make, we consider the combinations of characters:

1. Choose 2 capital letters, 1 digit, and 1 symbol: \(\binom{26}{2} \times 9 \times 5 \times 4! = 351,000\)
2. Choose 1 capital letter, 2 digits, and 1 symbol: \(26 \times \binom{9}{2} \times 5 \times 4! = 112,320\)
3. Choose 1 capital letter, 1 digit, and 2 symbols: \(26 \times 9 \times \binom{5}{2} \times 4! = 56,160\)

Summing these gives the total number of passwords: \(351,000 + 112,320 + 56,160 = 519,480\).

To solve this problem, we need to find the number of ways to distribute 12 different cakes between Alex and James such that both receive an odd number of cakes.

The possible distributions where both receive an odd number of cakes are:

• Alex receives 1 cake and James receives 11 cakes.
• Alex receives 3 cakes and James receives 9 cakes.
• Alex receives 5 cakes and James receives 7 cakes.
• Alex receives 7 cakes and James receives 5 cakes.
• Alex receives 9 cakes and James receives 3 cakes.
• Alex receives 11 cakes and James receives 1 cake.

For each distribution, we calculate the number of ways to choose the cakes for Alex, and the remaining cakes automatically go to James. This is given by the combination formula:

\(\binom{12}{1} + \binom{12}{3} + \binom{12}{5} + \binom{12}{7} + \binom{12}{9} + \binom{12}{11}\)

Calculating each term:

• \(\binom{12}{1} = 12\)
• \(\binom{12}{3} = 220\)
• \(\binom{12}{5} = 792\)
• \(\binom{12}{7} = 792\)
• \(\binom{12}{9} = 220\)
• \(\binom{12}{11} = 12\)

Summing these values gives:

\(12 + 220 + 792 + 792 + 220 + 12 = 2048\)

Therefore, the number of ways to distribute the cakes such that both receive an odd number is 2048.