(i) Consider the different combinations of players:

• 5 batsmen, 5 bowlers, 1 wicketkeeper: \(\binom{10}{5} \times \binom{9}{5} \times \binom{2}{1} = 63504\)
• 5 batsmen, 4 bowlers, 2 wicketkeepers: \(\binom{10}{5} \times \binom{9}{4} \times \binom{2}{2} = 31752\)
• 6 batsmen, 4 bowlers, 1 wicketkeeper: \(\binom{10}{6} \times \binom{9}{4} \times \binom{2}{1} = 52920\)

Total ways: \(63504 + 31752 + 52920 = 148176\)

(ii) The number of different arrangements of the presents is given by:

\(\frac{11!}{5!4!2!} = 6930\)

(iii) Consider omitting one type of present:

• Omit a pen: \(\frac{10!}{4!4!2!} = 3150\)
• Omit a diary: \(\frac{10!}{5!3!2!} = 2520\)
• Omit a notebook: \(\frac{10!}{5!4!} = 1260\)

Total arrangements: \(3150 + 2520 + 1260 = 6930\)