Answer:
The answer is a 100,000-choose-99 (a combination.)
[tex]P(100000, 99) = \displaystyle \left( \begin{array}{c}100000\cr 99\end{array}\right)[/tex].
Step-by-step explanation:
A combination [tex]C(n, r)[/tex] or equivalently [tex]\displaystyle \left(\begin{array}{c}n \cr r\end{array}\right)[/tex] gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements.
A permutation [tex]P(n, r)[/tex] also gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements. On top of that, it accounts for the order of the elements. Two elements in different order counts twice in a permutation, but only once in a combination.
The question emphasize that the council members are "co-equal." That implies that the order of the members don't really matter. Hence a combination with [tex]n = 100000[/tex] and [tex]r = 99[/tex] would be a more suitable choice.
