Answer: \(720\), \(480\), \(264\), and \(n=14\).
(a)(i) There are \(6\) digits available and \(5\) positions. No digit is repeated, so the number of possible 5-digit numbers is
\({}^6P_5=6\cdot5\cdot4\cdot3\cdot2=720.\)
(a)(ii) For the number to be odd, the final digit must be \(1,3,5\) or \(9\), giving \(4\) choices. The remaining four positions can then be filled by arranging \(4\) of the remaining \(5\) digits:
\(4\cdot{}^5P_4=4\cdot5\cdot4\cdot3\cdot2=480.\)
(a)(iii) The number must be odd and greater than \(60000\).
If it starts with \(6\) or \(8\), there are \(2\) choices for the first digit, \(4\) choices for the final odd digit, and then \({}^4P_3=24\) ways to fill the middle three positions. This gives
\(2\cdot4\cdot24=192.\)
If it starts with \(9\), the final digit must be \(1,3\) or \(5\), giving \(3\) choices, and again there are \(24\) ways to fill the middle three positions. This gives
\(3\cdot24=72.\)
So the total is
\(192+72=264.\)
(b) Use the formula
\(\binom nr=\frac{n!}{r!(n-r)!}.\)
The equation is
\(45\binom n4=(n+1)\binom{n+1}{5}.\)
Substitute the factorial forms:
\(45\cdot\frac{n!}{(n-4)!4!} =(n+1)\cdot\frac{(n+1)!}{((n+1)-5)!5!}.\)
After simplifying, this gives
\(45=\frac{(n+1)^2}{5}.\)
So
\((n+1)^2=225.\)
Since \(n\) is positive,
\(n+1=15.\)
Therefore
\(n=14.\)
