Answer: (a)(i) \(70\), (a)(ii) \(490\); (b) \(n=6\).

(a)(i) If all \(4\) doctors are on the committee, the remaining \(4\) members must be chosen from the \(5+3=8\) teachers and police officers.

So the number of committees is

\(\binom84=70.\)

(a)(ii) There are \(12\) people altogether, so the total number of committees is

\(\binom{12}{8}=495.\)

A committee with fewer than \(2\) teachers can only have exactly \(1\) teacher, because there are only \(7\) non-teachers.

The number of committees with exactly \(1\) teacher is

\(\binom51\binom77=5.\)

Therefore the number with at least \(2\) teachers is

\(495-5=490.\)

(b) Use the formula for permutations:

The equation becomes

\(\frac{n!}{(n-5)!} =6\frac{(n-1)!}{(n-5)!}.\)

Cancel the common factor:

\(n!=6(n-1)!.\)

Since \(n!=n(n-1)!\),

\(n=6.\)