Answer: (a)(i) \(5040\), (ii) \(2520\), (iii) \(4536\); (b) \(517\).

Use permutations when the order of the selected objects matters. Deal with restrictions by fixing or grouping the restricted objects first.

(a)(i) There are 10 choices for the first digit, then 9, then 8, then 7 choices for the remaining positions.

So the number of codes is

\(10\times9\times8\times7=5040\).

(a)(ii) For an odd code, the last digit must be one of \(1,3,5,7,9\), so there are 5 choices for the last digit.

After that, the first three positions can be filled in \(9\times8\times7\) ways.

Thus the number of odd codes is

\(5\times9\times8\times7=2520\).

(a)(iii) The total number of codes is \(5040\). A code is less than 1000 only when it begins with 0.

If the first digit is 0, the remaining three digits can be chosen in

\(9\times8\times7=504\)

ways.

Therefore the number of codes greater than 1000 is

\(5040-504=4536\).

(b) Since the family must not be separated, there are two cases.

If all 4 family members are chosen, we must choose 5 more people from the other 11:

\(\binom{11}{5}=462\).

If no family members are chosen, we choose all 9 people from the other 11:

\(\binom{11}{9}=55\).

So the total number of teams is

\(462+55=517\).