(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:

\(420 - 240 = 180\)