Answer: (a)(i) \(300\), (ii) \(252\), (iii) \(108\). (b) \(n=65\).

(a)(i) The first digit cannot be \(0\), so there are \(5\) choices for the first digit.

The remaining three positions can then be filled in \(5\times4\times3\) ways. Hence the number of 4-digit numbers is

\(5\times5\times4\times3=300.\)

(a)(ii) For an even number, the final digit is \(0,2,4,6\) or \(8\).

If the final digit is \(0\), there are

\(5\times4\times3=60\)

numbers.

If the final digit is one of \(2,4,6,8\), there are \(4\) choices for the final digit, \(4\) non-zero choices for the first digit, then \(4\) and \(3\) choices for the middle digits. This gives

\(4\times4\times4\times3=192.\)

Therefore the total is

\(60+192=252.\)

(a)(iii) A number divisible by \(5\) must end in \(0\) or \(5\).

If it ends in \(0\), there are

\(5\times4\times3=60\)

numbers.

If it ends in \(5\), the first digit has \(4\) choices, then the middle digits have \(4\) and \(3\) choices. This gives

\(4\times4\times3=48.\)

So the total is

\(60+48=108.\)

(b) Use

\({}^{n+1}C_{12}=\frac{(n+1)!}{12!(n-11)!}\)

and

\({}^nC_{10}=\frac{n!}{10!(n-10)!}.\)

The equation becomes

\((n+1)\frac{(n+1)!}{12!(n-11)!}=33(n-10)\frac{n!}{10!(n-10)!}.\)

Since \((n-10)!=(n-10)(n-11)!\), this simplifies to

\(\frac{(n+1)^2}{12\times11}=33.\)

Therefore

\((n+1)^2=33\times12\times11=4356.\)

So \(n+1=66\), and hence

\(n=65.\)