Question

Let $\left(\begin{array}{c}n\\ k\end{array}\right)$ denote ${}^n{C}_k$ and $\left[\begin{array}{c}n\\ k\end{array}\right]=\{\begin{array}{ccc}\left(\begin{array}{c}n\\ k\end{array}\right),& & \text{if}0\le k\le n\\ 0,& & \text{otherwise.}\end{array}$
If $A_k=\sum _{i=0}^9\left(\begin{array}{c}9\\ i\end{array}\right)\left[\begin{array}{c}12\\ 12-k+i\end{array}\right]+\sum _{i=0}^8\left(\begin{array}{c}8\\ i\end{array}\right)\left[\begin{array}{c}13\\ 13-k+i\end{array}\right]$ and $A_4-A_3=190p,$ then $p$ is equal to

Answer 1

$A_k=\sum _{i=0}^9{}^9{C}_i{}^{12}{C}_{12-k+i}+\sum _{k=0}^8{}^8{C}_i\cdot {}^{13}{C}_{13-k+i}$
$A_k=\sum _{k=0}^9{}^9{C}_i\cdot {}^{12}{C}_{k-i}+\sum _{i=0}^8{}^8{C}_i\cdot {}^{13}{C}_{k-i}$
$A_k=\text{Coeff of}{x}^k\text{in}(1+x{)}^9\cdot (1+x{)}^{12}+(1+x{)}^8\cdot (1+x{)}^{13}$
$A_k=2\cdot {}^{21}{C}_k$
$A_4-A_3=2\left[{}^{21}{C}_4{-}^{21}{C}_3\right]$
$=2[\frac{21\times 20\times 19\times 18}{24}-\frac{21\times 20\times 19}{6}]$
$=2\times 21\times 20\times 19[\frac{18}{24}-\frac{1}{6}]=190\times 49$
$\therefore p=49$