Answer: (a)(i) \(20160\), (ii) \(7200\), (iii) \(360\); (b) \(n=30\).

(a)(i) There are \(8\) choices for the first character, then \(7\), then \(6\), and so on for six positions.

(a)(ii) The first and last characters must be digits.

There are \(5\) choices for the first digit and \(4\) choices for the last digit. The four middle positions are then filled from the remaining \(6\) characters:

(a)(iii) The first three positions must be the three symbols, in any order. This can be done in \(3!\) ways.

The last three positions are chosen and arranged from the five digits:

\(3!\cdot{}^5P_3 =6(5\cdot4\cdot3)=360.\)

(b) The given statement is

\(\binom n5=6\binom{n-1}{4}.\)

Using factorials,

Cancel \((n-1)!\) and \((n-5)!\):

\(\frac{n}{5!}=\frac{6}{4!}.\)

Since \(5!=5\cdot4!\), this gives

\(\frac{n}{5}=6.\)

Therefore

\(n=30.\)