Answer: \(360\); \(60\); \(1287\); \(1266\); \(45\).

(a)(i) There are \(6\) choices for the first digit, then \(5\), then \(4\), then \(3\). Therefore the number is

\(6\times5\times4\times3=360.\)

(a)(ii) The number must be even and greater than \(6000\).

If it starts with \(6\), the final digit must be \(4\), and the two middle digits can be chosen in \(4\times3\) ways. This gives \(12\) numbers.

If it starts with \(7\) or \(9\), there are \(2\) choices for the first digit, \(2\) choices for the even final digit, and \(4\times3\) choices for the middle two digits. This gives

\(2\times2\times4\times3=48.\)

So the total number is

\(12+48=60.\)

(b)(i) There are \(13\) people in total, so the number of committees is

\(\binom{13}{5}=1287.\)

(b)(ii) It is easier to subtract the committees with no doctors. There are \(7\) non-doctors, so the number with no doctors is \(\binom75\). Hence the required number is

\(\binom{13}{5}-\binom75=1287-21=1266.\)

(b)(iii) If all \(3\) nurses are included, choose the other \(2\) members from the \(10\) doctors and dentists:

\(\binom{10}{2}=45.\)