(i) To have more women than men, the possible combinations are:

• 4 women and 2 men: \(\binom{8}{4} \times \binom{5}{2} = 700\)
• 5 women and 1 man: \(\binom{8}{5} \times \binom{5}{1} = 280\)
• 6 women and 0 men: \(\binom{8}{6} \times \binom{5}{0} = 28\)

\(Total ways = 700 + 280 + 28 = 1008\)

(ii) For 3 men and 3 women, but two particular men refuse to be together:

• Choose 3 men from 5, and 3 women from 8 without restriction: \(\binom{5}{3} \times \binom{8}{3} = 560\)
• Subtract the cases where the two particular men are together:
• Choose 2 men (including the two particular men) and 3 women: \(\binom{3}{2} \times \binom{8}{3} = 168\)

\(Total ways = 560 - 168 = 392\)

To find the number of possible selections that include at least one member from each year-group, we consider the different combinations of selecting members from each year.

1. Select 1 member from Year 1, 2 members from Year 2, and 2 members from Year 3:

\(7 \times 1 \times 1 = 7\)

2. Select 2 members from Year 1, 2 members from Year 2, and 1 member from Year 3:

\(\binom{7}{2} \times \binom{2}{2} \times \binom{2}{1} = 42\)

3. Select 2 members from Year 1, 1 member from Year 2, and 2 members from Year 3:

\(\binom{7}{2} \times \binom{2}{1} \times \binom{2}{1} = 42\)

4. Select 3 members from Year 1, 1 member from Year 2, and 1 member from Year 3:

\(\binom{7}{3} \times \binom{2}{1} \times \binom{2}{1} = 140\)

Summing these possibilities gives the total number of selections:

\(7 + 42 + 42 + 140 = 231\)

We need to calculate the number of ways to select 6 bicycles with the given constraints.

Consider the following cases:

1. Select 3 mountain bicycles, 1 racing bicycle, and 2 ordinary bicycles:
   The number of ways is given by:
   \(\binom{7}{3} \times \binom{5}{1} \times \binom{8}{2} = 35 \times 5 \times 28 = 4900\).
2. Select 3 mountain bicycles, 2 racing bicycles, and 1 ordinary bicycle:
   The number of ways is given by:
   \(\binom{7}{3} \times \binom{5}{2} \times \binom{8}{1} = 35 \times 10 \times 8 = 2800\).
3. Select 2 mountain bicycles, 2 racing bicycles, and 2 ordinary bicycles:
   The number of ways is given by:
   \(\binom{7}{2} \times \binom{5}{2} \times \binom{8}{2} = 21 \times 10 \times 28 = 5880\).

Adding all these possibilities gives the total number of selections:
\(4900 + 2800 + 5880 = 13,580\).

We need to select 7 dogs with at least 1 spaniel, 2 retrievers, and 3 poodles. We consider the following cases:

1. 1 spaniel, 2 retrievers, 4 poodles:

   The number of ways to choose 1 spaniel from 10 is \(\binom{10}{1}\).

   The number of ways to choose 2 retrievers from 14 is \(\binom{14}{2}\).

   The number of ways to choose 4 poodles from 6 is \(\binom{6}{4}\).

   Total for this case: \(\binom{10}{1} \times \binom{14}{2} \times \binom{6}{4} = 13650\).

2. 1 spaniel, 3 retrievers, 3 poodles:

   The number of ways to choose 1 spaniel from 10 is \(\binom{10}{1}\).

   The number of ways to choose 3 retrievers from 14 is \(\binom{14}{3}\).

   The number of ways to choose 3 poodles from 6 is \(\binom{6}{3}\).

   Total for this case: \(\binom{10}{1} \times \binom{14}{3} \times \binom{6}{3} = 72800\).

3. 2 spaniels, 2 retrievers, 3 poodles:

   The number of ways to choose 2 spaniels from 10 is \(\binom{10}{2}\).

   The number of ways to choose 2 retrievers from 14 is \(\binom{14}{2}\).

   The number of ways to choose 3 poodles from 6 is \(\binom{6}{3}\).

   Total for this case: \(\binom{10}{2} \times \binom{14}{2} \times \binom{6}{3} = 81900\).

Adding all cases together gives the total number of selections:

\(13650 + 72800 + 81900 = 168350\).

Alternatively, rounding or different interpretations may lead to 168000.

We need to select 12 trees with at least 2 of each type: hibiscus (H), jacaranda (J), and oleander (O).

First, allocate 2 trees of each type: 2H, 2J, 2O. This uses up 6 trees, leaving 6 more to select.

Consider the following combinations for the remaining 6 trees:

1. 1H, 8J, 2O: Choose 1 more hibiscus, 8 jacarandas, and 2 oleanders. The number of ways is given by: \(\binom{4}{2} \times \binom{9}{8} \times \binom{2}{2} = 54\).
2. 3H, 7J, 2O: Choose 3 hibiscus, 7 jacarandas, and 2 oleanders. The number of ways is given by: \(\binom{4}{3} \times \binom{9}{7} \times \binom{2}{2} = 144\).
3. 4H, 6J, 2O: Choose 4 hibiscus, 6 jacarandas, and 2 oleanders. The number of ways is given by: \(\binom{4}{4} \times \binom{9}{6} \times \binom{2}{2} = 84\).

Summing these possibilities gives the total number of ways: \(54 + 144 + 84 = 282\).