The value of ${ }^{10} C_{1}+{ }^{10} C_{2}+{ }^{10} C_{3}+\ldots+{ }^{10} C_{9}$ is

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

If $^nC_0 + \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 + \dots + \frac{1}{n+1} ^nC_n = \frac{1023}{10}$, then $n =$ ?

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:

If $n$ is an integer greater than $1$,then $a - ^nC_1(a - 1) + ^nC_2(a - 2) + \dots + (-1)^n(a - n) = $

If $c_0, c_1, c_2, \ldots, c_n$ denote the coefficients in the expansion of $(1+x)^n$, then the value of $c_1 + 2c_2 + 3c_3 + \ldots + nc_n$ is