We need to find the number of ways to form a team of 7 people with exactly 1 swimmer, at least 4 cyclists, and at most 2 runners.

Consider the following scenarios:

1. 1 swimmer, 4 cyclists, 2 runners:
   The number of ways to choose 1 swimmer from 6 is \(\binom{6}{1}\).
   The number of ways to choose 4 cyclists from 8 is \(\binom{8}{4}\).
   The number of ways to choose 2 runners from 11 is \(\binom{11}{2}\).
   Total for this scenario: \(\binom{6}{1} \times \binom{8}{4} \times \binom{11}{2} = 6 \times 70 \times 55 = 23100\).
2. 1 swimmer, 5 cyclists, 1 runner:
   The number of ways to choose 1 swimmer from 6 is \(\binom{6}{1}\).
   The number of ways to choose 5 cyclists from 8 is \(\binom{8}{5}\).
   The number of ways to choose 1 runner from 11 is \(\binom{11}{1}\).
   Total for this scenario: \(\binom{6}{1} \times \binom{8}{5} \times \binom{11}{1} = 6 \times 56 \times 11 = 3696\).
3. 1 swimmer, 6 cyclists, 0 runners:
   The number of ways to choose 1 swimmer from 6 is \(\binom{6}{1}\).
   The number of ways to choose 6 cyclists from 8 is \(\binom{8}{6}\).
   The number of ways to choose 0 runners from 11 is \(\binom{11}{0}\).
   Total for this scenario: \(\binom{6}{1} \times \binom{8}{6} \times \binom{11}{0} = 6 \times 28 \times 1 = 168\).

Adding these scenarios together gives the total number of ways:
\(23100 + 3696 + 168 = 26964\).

First, calculate the total number of ways to choose 5 people from 10 (4 men + 6 women):

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

Next, calculate the number of ways William and Mary are both in the committee. If William and Mary are both chosen, we need to choose 3 more people from the remaining 8 people:

\(\binom{8}{3} = 56\)

Therefore, the number of committees where William and Mary are not together is:

\(252 - 56 = 196\)

(i) To find the number of different teams, calculate the combinations for each group:

Choose 5 batsmen from 7: \(\binom{7}{5}\)

Choose 4 bowlers from 5: \(\binom{5}{4}\)

Choose 1 all-rounder from 2: \(\binom{2}{1}\)

Choose 1 wicket-keeper from 2: \(\binom{2}{1}\)

The total number of combinations is:

\(\binom{7}{5} \times \binom{5}{4} \times \binom{2}{1} \times \binom{2}{1} = 21 \times 5 \times 2 \times 2 = 420\)

(ii) If one particular batsman refuses to be in the team when one particular bowler is in the team, calculate the number of teams where both are in the team and subtract from the total:

Choose 4 batsmen from the remaining 6: \(\binom{6}{4}\)

Choose 3 bowlers from the remaining 4: \(\binom{4}{3}\)

Choose 1 all-rounder from 2: \(\binom{2}{1}\)

Choose 1 wicket-keeper from 2: \(\binom{2}{1}\)

The number of teams with both in the team is:

\(\binom{6}{4} \times \binom{4}{3} \times 2 \times 2 = 15 \times 4 \times 2 \times 2 = 240\)

Subtract from the total number of teams:

\(420 - 240 = 180\)

We need to choose a team of 6 people with at least 1 swimmer, 1 athlete, and 1 cyclist, and more athletes than cyclists. Let's consider the possible distributions:

1. 1 swimmer, 3 athletes, 2 cyclists:

   \(egin{align*} 5 \times \binom{7}{3} \times \binom{4}{2} &= 5 \times 35 \times 6 = 1050 ext{ ways} div>\)

2. 1 swimmer, 4 athletes, 1 cyclist:

   \(egin{align*} 5 \times \binom{7}{4} \times 4 &= 5 \times 35 \times 4 = 700 ext{ ways} div>\)

3. 2 swimmers, 3 athletes, 1 cyclist:

   \(egin{align*} \binom{5}{2} \times \binom{7}{3} \times 4 &= 10 \times 35 \times 4 = 1400 ext{ ways} div>\)

4. 3 swimmers, 2 athletes, 1 cyclist:

   \(egin{align*} \binom{5}{3} \times \binom{7}{2} \times 4 &= 10 \times 21 \times 4 = 840 ext{ ways} div>\)

Adding all these possibilities gives the total number of ways:

\(egin{align*} 1050 + 700 + 1400 + 840 = 3990 ext{ ways} div>\)

We need to find the number of ways to select 7 flowers with at least 2 of each type (roses, tulips, daffodils) from the available stock: 6 roses, 5 tulips, and 4 daffodils.

Consider the possible distributions of flowers:

1. 2 roses, 2 tulips, 3 daffodils: \(\binom{6}{2} \times \binom{5}{2} \times \binom{4}{3} = 600\)
2. 2 roses, 3 tulips, 2 daffodils: \(\binom{6}{2} \times \binom{5}{3} \times \binom{4}{2} = 900\)
3. 3 roses, 2 tulips, 2 daffodils: \(\binom{6}{3} \times \binom{5}{2} \times \binom{4}{2} = 1200\)

Summing these possibilities gives the total number of different bunches:

\(600 + 900 + 1200 = 2700\)