Answer: (a)(i) \(17640\); (a)(ii) \(4680\); (a)(iii) \(3960\); (b) \(135\).

Answer: (a)(i) \(17640\); (a)(ii) \(4680\); (a)(iii) \(3960\); (b) \(135\).

(a)(i) The first digit has \(7\) choices, since it cannot be \(0\). After that, \(5\) more digits are chosen and arranged from the remaining \(7\) digits.

So the number of possible numbers is

\(7\times {}^7P_5=7\times2520=17640.\)

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

If it ends in \(5\), the first digit has \(6\) choices, and the middle four positions can be filled in \({}^6P_4\) ways. This gives

\(6\times{}^6P_4=2160.\)

If it ends in \(0\), the first digit has \(7\) choices, and the middle four positions can be filled in \({}^6P_4\) ways. This gives

\(7\times{}^6P_4=2520.\)

Hence the total is

\(2160+2520=4680.\)

(a)(iii) For numbers starting with \(85\), \(86\), \(87\), or \(89\), the remaining four digits can be chosen and arranged in \({}^6P_4=360\) ways. This gives

\(4\times360=1440.\)

For numbers starting with \(9\), the remaining five digits can be chosen and arranged in

\({}^7P_5=2520\)

ways. Therefore the total number greater than \(850000\) is

\(1440+2520=3960.\)

(b) The three brothers must either all be chosen or all be left out.

If all three brothers are chosen, choose the remaining \(5\) people from the other \(9\):

\(\binom95=126.\)

If none of the brothers are chosen, choose all \(8\) people from the other \(9\):

\(\binom98=9.\)

So the total number of teams is

\(126+9=135.\)