Answer: (a) \(2522520\); (b)(i) \(136080\); (ii) \(15120\); (iii) \(38640\).
Use combinations when order is not important and permutations when order is important. Deal with restrictions by choosing the restricted items first, then choosing the remaining items.
(a) The groups have different sizes, so the number of ways is
\(\frac{14!}{2!3!4!5!}=2522520.\)
(b)(i) For a 6-digit number, the first digit cannot be 0. So there are 9 choices for the first digit.
After that, there are 9, 8, 7, 6 and 5 choices for the remaining places.
Thus the number of possibilities is
\(9\times9\times8\times7\times6\times5=136080.\)
(b)(ii) A number divisible by 10 must end in 0. The first digit then has 9 choices, and the four middle positions have
\(8\times7\times6\times5\)
choices. So the number of possibilities is
\(9\times8\times7\times6\times5=15120.\)
(b)(iii) The first digit must be one of \(5,6,7,8,9\), and the final digit must be even.
If the final digit is \(0\), there are 5 choices for the first digit and then \(8P4\) choices for the middle four digits:
\(5\times(8\times7\times6\times5)=8400.\)
If the final digit is \(2\) or \(4\), there are 2 choices for the final digit, 5 choices for the first digit, and \(8P4\) choices for the middle four digits:
\(2\times5\times(8\times7\times6\times5)=16800.\)
If the final digit is \(6\) or \(8\), there are 2 choices for the final digit, but only 4 choices for the first digit because it cannot equal the final digit:
\(2\times4\times(8\times7\times6\times5)=13440.\)
Therefore the total is
\(8400+16800+13440=38640.\)
