Answer: (a)(i) \(35700\); (a)(ii) \(32400\); (b)(ii) \(n=7\).

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) There are two cases.

If neither Ann nor Bo is on the committee, all 6 members are chosen from the remaining 18 members:

\(\binom{18}{6}=18564.\)

If exactly one of Ann and Bo is on the committee, choose which one in 2 ways, then choose the other 5 committee members from the remaining 18 members:

\(2\binom{18}{5}=2(8568)=17136.\)

Therefore the total number of committees is

\(18564+17136=35700.\)

(a)(ii) The first 4 characters must be letters. These can be arranged in

\({}^6P_4=6\cdot5\cdot4\cdot3=360\)

ways. The final 2 characters must be digits. These can be arranged in

\({}^{10}P_2=10\cdot9=90\)

ways. Hence the number of passcodes is

\(360\cdot90=32400.\)

(b)(i) Use the formula for combinations:

\((n-3)\binom n3 =(n-3)\frac{n!}{3!(n-3)!}.\)

Since \((n-3)!=(n-3)(n-4)!\), this simplifies to

\(\frac{n!}{3!(n-4)!}.\)

Also,

\(4\binom n4=4\frac{n!}{4!(n-4)!} =\frac{n!}{3!(n-4)!}.\)

Therefore

\((n-3)\binom n3=4\binom n4.\)

(b)(ii) Start with

\(\binom n3=5n.\)

Then

\(\frac{n(n-1)(n-2)}{6}=5n.\)

Since \(n\gt 3\), divide by \(n\):

\((n-1)(n-2)=30.\)

Expanding gives

\(n^2-3n+2=30,\)

so

\(n^2-3n-28=0.\)

Factorising,

\((n-7)(n+4)=0.\)

Since \(n\gt 3\),

\(n=7.\)