Answer: (a)(i) \(6435\); (ii) \(495\); (b) \(n=27\).

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)(i) The number of possible teams is

\(\binom{15}{8}=6435.\)

(a)(ii) If all four family members are included, choose the remaining 4 people from the other 11:

\(\binom{11}{4}=330.\)

If no family members are included, choose all 8 people from the other 11:

\(\binom{11}{8}=165.\)

So the total number of possible teams is

\(330+165=495.\)

(b) Use

The equation becomes

\((n+9)\frac{n!}{(n-10)!}=(n^2+243)\frac{(n-1)!}{(n-10)!}.\)

Cancel the common factor \(\frac{(n-1)!}{(n-10)!}\):

\(n(n+9)=n^2+243.\)

So

\(n^2+9n=n^2+243.\)

Hence

\(9n=243,\)

and therefore

\(n=27.\)