# Solve this

Question:

Let $\left(\begin{array}{l}\mathrm{n} \\ \mathrm{k}\end{array}\right)$ denotes ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{k}}$ and $\left[\begin{array}{l}\mathrm{n} \\ \mathrm{k}\end{array}\right]=\left\{\begin{array}{cl}\left(\begin{array}{l}\mathrm{n} \\ \mathrm{k}\end{array}\right), & \text { if } 0 \leq \mathrm{k} \leq \mathrm{n} \\ 0, & \text { otherwise }\end{array}\right.$

If $A_{k}=\sum_{i=0}^{9}\left(\begin{array}{l}9 \\ i\end{array}\right)\left[\begin{array}{c}12 \\ 12-k+i\end{array}\right]+\sum_{i=0}^{8}\left(\begin{array}{c}8 \\ i\end{array}\right)\left[\begin{array}{c}13 \\ 13-k+i\end{array}\right]$

and $\mathrm{A}_{4}-\mathrm{A}_{3}=190 \mathrm{p}$, then $\mathrm{p}$ is equal to :

Solution:

$A_{k}=\sum_{i=0}^{9}{ }^{9} C_{i}{ }^{12} C_{k-i}+\sum_{i=0}^{8}{ }^{8} C_{i}{ }^{13} C_{k-i}$

$\mathrm{A}_{\mathrm{k}}={ }^{21} \mathrm{C}_{\mathrm{k}}+{ }^{21} \mathrm{C}_{\mathrm{k}}=2 \cdot{ }^{21} \mathrm{C}_{\mathrm{k}}$

$\mathrm{A}_{4}-\mathrm{A}_{3}=2\left({ }^{21} \mathrm{C}_{4}-{ }^{21} \mathrm{C}_{3}\right)=2(5985-1330)$

$190 \mathrm{p}=2(5985-1330) \Rightarrow \mathrm{p}=49$