If $\left({}^{30}C_1\right)^2 + 2\left({}^{30}C_2\right)^2 + 3\left({}^{30}C_3\right)^2 + \ldots + 30\left({}^{30}C_{30}\right)^2 = \frac{\alpha 60!}{(30!)^2}$ then $\alpha$ is equal to:
\( 10 \)
\( 15 \)
\( 30 \)
\( 60 \)