Answer: (a)(i) \(720\), (ii) \(600\), (iii) \(168\); (b) \(n=20\).
(a)(i) A 5-digit number is made by choosing and arranging 5 of the 6 digits.
So the number of possible numbers is
\({}^6P_5=6\times5\times4\times3\times2=720.\)
(a)(ii) A number is divisible by \(5\) only if its last digit is \(5\). If the last digit is fixed as \(5\), the other 4 positions can be filled from the remaining 5 digits:
\({}^5P_4=5\times4\times3\times2=120.\)
Hence the number not divisible by \(5\) is
\(720-120=600.\)
(a)(iii) For the number to be even, the last digit must be \(2\) or \(8\).
If the last digit is \(2\), the first digit must be one of \(3,5,7,8\), so there are \(4\) choices for the first digit. The middle 3 positions can then be filled in
\(4\times3\times2=24\)
ways. This gives
\(4\times24=96\)
numbers.
If the last digit is \(8\), the first digit must be one of \(3,5,7\), so there are \(3\) choices for the first digit. The middle 3 positions can again be filled in \(24\) ways. This gives
\(3\times24=72\)
numbers.
Therefore the total number is
\(96+72=168.\)
(b) Since
\({}^nC_3=6\,{}^nC_2,\)
we have
\(\frac{n(n-1)(n-2)}{3!}=6\cdot\frac{n(n-1)}{2!}.\)
For \(n\gt2\), divide by \(n(n-1)\):
\(\frac{n-2}{6}=3.\)
So
\(n-2=18,\)
and hence
\(n=20.\)
