If  $\sum _{k=1}^{31}{(}^{31}{C}_k){(}^{31}{C}_{k-1})-\sum _{k=1}^{30}{(}^{30}{C}_k){(}^{30}{C}_{k-1})=\frac{\alpha (60!)}{(30!)(31!)},$  Where  $\alpha \in R,$  then the value of  $16\alpha$  is equal to k, where the difference between maximum and minimum digits of k is _______

$$\begin{gathered} {\sum _{R=1}^{31}}^{31}{C}_R{.}^{31}{C}_{R-1} \\ {=}^{31}{C}_1{.}^{31}{C}_0{+}^{31}{C}_2{.}^{31}{C}_1+\mathrm{........}{+}^{31}{C}_{31}{.}^{31}{C}_{30} \\ {=}^{31}{C}_0{.}^{31}{C}_{30}{+}^{31}{C}_1{.}^{31}{C}_{29}+\mathrm{..........}{+}^{31}{C}_{30}{.}^{31}{C}_0{=}^{62}{C}_{30} \end{gathered}$$

Similarly,  $\sum _{R=1}^{30}{(}^{30}{C}_R{.}^{30}{C}_{R-1}){=}^{60}{C}_{29}$

$$\begin{gathered} { }^{62}{C}_{30}{-}^{60}{C}_{29}=\frac{62!}{30!32!}-\frac{60!}{29!31!} \\ =\frac{60!}{29!31!}\{\frac{62.61}{30.32}-1\}=\frac{60!}{30!31!}\left(\frac{2822}{32}\right) \end{gathered}$$

$\therefore 16\alpha =16\times \frac{2822}{32}=1411$  maximum digit = 4 minimum digit = 1