Answer: (a)(i) \(136080\), (ii) \(28560\); (b)(i) \(346104\), (ii) \(18\), (iii) \(334664\).

Use combinations when order is not important and permutations when order is important. Deal with restrictions by choosing the restricted items first, then choosing the remaining items.

(a)(i) The first digit cannot be 0, so there are 9 choices for the first digit.

After that, there are 9, 8, 7, 6 and 5 choices for the remaining positions.

So the number of 6-digit numbers is

\(9\times9\times8\times7\times6\times5=136080\).

(a)(ii) A number divisible by 5 must end in 0 or 5.

If it ends in 0, the first digit has 9 choices, and the middle four positions have \(8\times7\times6\times5\) choices.

This gives

\(9\times8\times7\times6\times5=15120\).

If it ends in 5, the first digit cannot be 0 or 5, so it has 8 choices. The middle four positions then have \(8\times7\times6\times5\) choices.

This gives

\(8\times8\times7\times6\times5=13440\).

Therefore the total number divisible by 5 is

\(15120+13440=28560\).

(b)(i) There are \(6+10+8=24\) people in total.

The number of committees of 7 is

\(\binom{24}{7}=346104\).

(b)(ii) If all 6 doctors are on the committee, one more person must be chosen from the 18 non-doctors.

This can be done in

\(\binom{18}{1}=18\)

ways.

(b)(iii) It is easier to subtract the committees with no dentists from all committees.

The number with no dentists is the number of committees of 7 chosen from the 16 doctors and nurses:

\(\binom{16}{7}=11440\).

Therefore the number with at least one dentist is

\(\binom{24}{7}-\binom{16}{7}=346104-11440=334664\).