Two players A and B play a series of games of badminton. The...

Player $A$ wins the series if $A$ wins $5$ games before $B$ wins $5$ games. The series can last for a minimum of $5$ games and a maximum of $9$ games.

If player $A$ wins the series in exactly $n$ games, then $A$ must win the $n$-th game, and $A$ must win exactly $4$ out of the first $n-1$ games. The number of ways this can happen is given by $^{n-1}C_{4}$.

The possible values for $n$ are $5, 6, 7, 8,$ and $9$.

Total number of ways for $A$ to win the series is the sum of the number of ways $A$ can win in $5, 6, 7, 8,$ or $9$ games:

Total ways $= ^{4}C_{4} + ^{5}C_{4} + ^{6}C_{4} + ^{7}C_{4} + ^{8}C_{4}$

Using the property $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$, we can simplify the sum:
$^{4}C_{4} = ^{5}C_{5}$
$^{5}C_{5} + ^{5}C_{4} = ^{6}C_{5}$
$^{6}C_{5} + ^{6}C_{4} = ^{7}C_{5}$
$^{7}C_{5} + ^{7}C_{4} = ^{8}C_{5}$
$^{8}C_{5} + ^{8}C_{4} = ^{9}C_{5}$

Therefore, the total number of ways is:
$^{9}C_{5} = \dfrac{9!}{5!4!} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

Answer: $126$