The sum $\sum_{r=1}^{9} \frac{10!}{r!(10-r)!}$ is equal to:
(1) $2^{10} - 2$
(2) $2^{10} - 1$
(3) $2^9$
(4) $2^{10}$
The sum $\sum_{r=1}^{9} \frac{10!}{r!(10-r)!}$ is equal to:\\
(1) $2^{10} - 2$\\
(2) $2^{10} - 1$\\
(3) $2^9$\\
(4) $2^{10}$