Answer: (a)(i) \(360\), (ii) \(60\), (iii) \(120\); (b) \(n=25\).

(a)(i) We choose and arrange 4 digits from 6 different digits, with no repetition.

So the number of 4-digit numbers is

\({}^6P_4=6\times5\times4\times3=360.\)

(a)(ii) A number is divisible by 5 if its final digit is 5.

Fix the final digit as 5. The first three positions can then be filled by choosing and arranging 3 of the remaining 5 digits:

\({}^5P_3=5\times4\times3=60.\)

(a)(iii) The number must be odd and greater than \(7000\).

If the number starts with \(8\), the final digit can be \(3\), \(5\), \(7\) or \(9\). This gives

\(4\times{}^4P_2=4\times12=48\)

numbers.

If the number starts with \(7\), the final digit can be \(3\), \(5\) or \(9\). This gives

\(3\times{}^4P_2=3\times12=36\)

numbers.

If the number starts with \(9\), the final digit can be \(3\), \(5\) or \(7\). This also gives

\(3\times{}^4P_2=36\)

numbers.

Therefore the total is

\(48+36+36=120.\)

(b) We are given

\({}^nC_3=92n.\)

So

\(\frac{n!}{(n-3)!3!}=92n.\)

This gives

\(\frac{n(n-1)(n-2)}{6}=92n.\)

Since \(n\gt 0\), divide by \(n\):

\((n-1)(n-2)=552.\)

Expand and rearrange:

\(n^2-3n+2=552,\)

so

\(n^2-3n-550=0.\)

Factorising,

\((n-25)(n+22)=0.\)

Since \(n\) must be positive,

\(n=25.\)