(i) There are 11 questions in total (8 from Part A and 3 from Part B). The number of ways to choose 6 questions from 11 is given by the combination formula:

\(\binom{11}{6} = 462\)

(ii) Candidates must choose at least 4 questions from Part A. We consider the following cases:

• 4 questions from Part A and 2 from Part B: \(\binom{8}{4} \times \binom{3}{2} = 70 \times 3 = 210\)
• 5 questions from Part A and 1 from Part B: \(\binom{8}{5} \times \binom{3}{1} = 56 \times 3 = 168\)
• 6 questions from Part A: \(\binom{8}{6} = 28\)

Adding these gives: \(210 + 168 + 28 = 406\)

(iii) Candidates must either choose both question 1 and question 2 in Part A, or choose neither. Consider the cases:

• Choose both question 1 and 2, then choose 4 more from the remaining 9 questions: \(\binom{9}{4} = 126\)
• Choose neither question 1 nor 2, then choose 6 from the remaining 9 questions: \(\binom{9}{6} = 84\)

Adding these gives: \(126 + 84 = 210\)

(i) If both boys are in the team, we need to choose 2 more people from the remaining 7 people (3 girls and 4 adults). The number of ways to choose 2 people from 7 is given by the combination formula: \(\binom{7}{2} = 21\).

(ii) If the adults are either all in the team or all not in the team, we consider two cases:

• All adults in the team: This is not possible since the team size is 4 and there are 4 adults.
• All adults not in the team: We choose all 4 members from the 5 non-adults (2 boys and 3 girls). The number of ways is \(\binom{5}{4} = 5\).

Thus, the total number of ways is \(1 + 5 = 6\).

(iii) If at least 2 girls are in the team, we consider two cases:

• 2 girls in the team: Choose 2 girls from 3 and 2 others from the remaining 6 people (2 boys and 4 adults). The number of ways is \(\binom{3}{2} \times \binom{6}{2} = 3 \times 15 = 45\).
• 3 girls in the team: Choose 3 girls from 3 and 1 other from the remaining 6 people. The number of ways is \(\binom{3}{3} \times \binom{6}{1} = 1 \times 6 = 6\).

Thus, the total number of ways is \(45 + 6 = 51\).

(ii) To find the number of ways to choose 6 images with 2 from each folder, we calculate:

\(\binom{3}{2} \times \binom{4}{2} \times \binom{8}{2} = 3 \times 6 \times 28 = 504\)

(iii) We need to consider different combinations of choosing images such that there are at least 3 from the friends folder and at least 1 from each of the other two folders. The possible distributions are:

1. 3 from Friends, 1 from Family, 2 from Holiday:

\(\binom{8}{3} \times \binom{3}{1} \times \binom{4}{2} = 56 \times 3 \times 6 = 1008\)

2. 3 from Friends, 2 from Family, 1 from Holiday:

\(\binom{8}{3} \times \binom{3}{2} \times \binom{4}{1} = 56 \times 3 \times 4 = 672\)

3. 4 from Friends, 1 from Family, 1 from Holiday:

\(\binom{8}{4} \times \binom{3}{1} \times \binom{4}{1} = 70 \times 3 \times 4 = 840\)

Summing these possibilities gives the total number of ways:

\(1008 + 672 + 840 = 2520\)

Let the number of geraniums be denoted by \(G\), roses by \(R\), and lilies by \(L\). We have the following conditions:

1. \(G + R + L = 25\)

2. \(G \geq 11\)

3. \(R = L\)

From condition 3, we have \(R = L\), so \(G + 2R = 25\).

We consider the possible values for \(G\):

1. \(G = 11\): Then \(11 + 2R = 25\) gives \(R = 7\) and \(L = 7\). The number of ways to choose the flowers is \(\binom{15}{11} \times \binom{10}{7} \times \binom{8}{7} = 1310400\).
2. \(G = 13\): Then \(13 + 2R = 25\) gives \(R = 6\) and \(L = 6\). The number of ways to choose the flowers is \(\binom{15}{13} \times \binom{10}{6} \times \binom{8}{6} = 617400\).
3. \(G = 15\): Then \(15 + 2R = 25\) gives \(R = 5\) and \(L = 5\). The number of ways to choose the flowers is \(\binom{15}{15} \times \binom{10}{5} \times \binom{8}{5} = 14112\).

Adding these possibilities gives the total number of selections:

\(1310400 + 617400 + 14112 = 1941912\).

The word HAPPINESS consists of the letters H, A, P, P, I, N, E, S, S. We need to select 4 letters with no Ps and either one or two Ss.

First, remove the Ps, leaving us with the letters H, A, I, N, E, S, S.

Case 1: One S

Select 1 S and 3 other letters from H, A, I, N, E. This can be done in \(\binom{5}{3} = 10\) ways.

Case 2: Two Ss

Select 2 Ss and 2 other letters from H, A, I, N, E. This can be done in \(\binom{5}{2} = 10\) ways.

\(Total number of selections = 10 + 10 = 20.\)

Alternatively, select 3 letters from H, A, I, N, E, S (6 letters total) and ensure at least one S is included. This can be done in \(\binom{6}{3} = 20\) ways.