If $$\sum_{k=1}^{31} \left(^{31}C_{k}\right) \left(^{31}C_{k-1} \right) - \sum_{k=1}^{30} \left(^{30}C_{k}\right) \left(^{30}C_{k-1} \right)= \frac{\alpha(60!)}{(30!)(31!)}$$, where $$\alpha \in R$$, then the value of $$16\alpha$$ is equal to

We need to evaluate $$\sum_{k=1}^{31} \binom{31}{k}\binom{31}{k-1} - \sum_{k=1}^{30} \binom{30}{k}\binom{30}{k-1}$$ and express it as $$\frac{\alpha \cdot 60!}{30! \cdot 31!}$$, then find $$16\alpha$$.

Noting that $$\binom{31}{k-1} = \binom{31}{32-k}$$ and applying the Vandermonde identity $$\sum_{k} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$$ with $$m=n=31$$ and $$r=32$$, we obtain

$$\sum_{k=1}^{31} \binom{31}{k}\binom{31}{32-k} = \binom{62}{32}\,. $$

Similarly, since $$\binom{30}{k-1} = \binom{30}{31-k}$$ and using Vandermonde with $$m=n=30$$ and $$r=31$$, we get

$$\sum_{k=1}^{30} \binom{30}{k}\binom{30}{31-k} = \binom{60}{31}\,. $$

Therefore the left-hand side simplifies to

$$\binom{62}{32} - \binom{60}{31}\,. $$

We set this equal to $$\frac{\alpha \cdot 60!}{30! \cdot 31!}$$, so that

$$\alpha = \bigl(\binom{62}{32} - \binom{60}{31}\bigr)\frac{30! \cdot 31!}{60!}\,.$$

For the first term,

$$\binom{62}{32}\times\frac{30! \cdot 31!}{60!} =\frac{62!}{32! \cdot 30!}\times\frac{30! \cdot 31!}{60!} =\frac{62! \cdot 31!}{32! \cdot 60!} =\frac{62 \cdot 61}{32} =\frac{3782}{32} =\frac{1891}{16}\,. $$

For the second term,

$$\binom{60}{31}\times\frac{30! \cdot 31!}{60!} =\frac{60!}{31! \cdot 29!}\times\frac{30! \cdot 31!}{60!} =\frac{30!}{29!} =30\,. $$

Hence

$$\alpha = \frac{1891}{16} - 30 = \frac{1891 - 480}{16} = \frac{1411}{16},$$

so

$$16\alpha = 1411\,. $$

Therefore, the correct answer is Option A: 1411.