Suppose $\sum_{r=0}^{2023} r \cdot ^{2023}C_r = 2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $............$

$\frac{C_1}{C_0} + 2\frac{C_2}{C_1} + 3\frac{C_3}{C_2} + \dots + 15\frac{C_{15}}{C_{14}} = $

If $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^n$,where $n$ is a power of $2$,find $k$. Specifically,if $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^4$,then $k=$

$A$ binary sequence is an array of $0$'s and $1$'s. The number of $n$-digit binary sequences which contain an even number of $0$'s is

$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

Let $\binom{n}{k}$ denote ${}^{n}C_{k}$ and $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{if } 0 \leq k \leq n \\ 0, & \text{otherwise} \end{cases}$. If $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ and $A_{4}-A_{3}=190p$,then $p$ is equal to: