Answer: (a)(i) \(154440\); (a)(ii) \(124200\); (b) \(n=47\).

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 13 characters, and a password has 5 characters with no repetition. Therefore the number of possible passwords is

\({}^{13}P_5=13\cdot12\cdot11\cdot10\cdot9.\)

So

\({}^{13}P_5=154440.\)

(a)(ii) Count the passwords with at least one symbol by subtracting the passwords with no symbols from the total.

There are 10 non-symbol characters, so the number of passwords with no symbols is

\({}^{10}P_5=10\cdot9\cdot8\cdot7\cdot6=30240.\)

Hence the number with at least one symbol is

\(154440-30240=124200.\)

(b) Start with

\(16\binom n{12}=(n-10)\binom{n+1}{11}.\)

Using factorial notation,

\(16\frac{n!}{12!(n-12)!} =(n-10)\frac{(n+1)!}{11!(n-10)!}.\)

On the right-hand side, \((n-10)!=(n-10)(n-11)!\), so the factor \(n-10\) cancels. Thus

\(16\frac{n!}{12!(n-12)!} =\frac{(n+1)!}{11!(n-11)!}.\)

Since \(12!=12\cdot11!\), the left-hand side becomes

\(\frac{4}{3}\cdot\frac{n!}{11!(n-12)!}.\)

The right-hand side is

\(\frac{(n+1)n!}{11!(n-11)(n-12)!}.\)

Cancel the common factor \(\frac{n!}{11!(n-12)!}\):

\(\frac43=\frac{n+1}{n-11}.\)

So

\(4(n-11)=3(n+1).\)

Therefore

\(4n-44=3n+3,\)

which gives

\(n=47.\)