If $\left(\frac{1}{{ }^{15} \mathrm{C}_0}+\frac{1}{{ }^{15} \mathrm{C}_1}\right)\left(\frac{1}{{ }^{15} \mathrm{C}_1}+\frac{1}{{ }^{15} \mathrm{C}_2}\right) \ldots\left(\frac{1}{{ }^{15} \mathrm{C}_{12}}+\frac{1}{{ }^{15} \mathrm{C}_{13}}\right)=\frac{\alpha^{13}}{{ }^{14} \mathrm{C}_0{ }^{14} \mathrm{C}_1 \cdots{ }^{14} \mathrm{C}_{12}}$, then $30 \alpha$ is equal to $\_\_\_\_$ .

The product of the last two digits of $(1919)^{1919}$ is

The sum of the series $2 \times 1 \times{ }^{20} \mathrm{C}_4-3 \times 2 \times{ }^{20} \mathrm{C}_5+4 \times 3 \times{ }^{20} \mathrm{C}_6-5 \times 4 \times{ }^{20} \mathrm{C}_7+\cdots \cdots+18 \times 17 \times{ }^{20} \mathrm{C}_{20}$, is equal to ____________.