If $C_0, C_1, C_2, \ldots, C_{10}$ represent the binomial coefficients in the expansion of $(1+x)^{10}$,then $C_0 C_6+C_1 C_7+C_2 C_8+C_3 C_9+C_4 C_{10}=$

$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$

The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

Statement $-1$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} = (n+2) 2^{n-1}$
Statement $-2$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} x^r = (1+x)^n + nx(1+x)^{n-1}$

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:

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: