If $^{n} C _{9}=\,\,^{n} C _{8},$ find $^{n} C _{17}$

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

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

The set $S = \left\{ {1,2,3, \ldots ,12} \right\}$ is to be partitioned into three sets $A,\,B,\, C$ of equal size . Thus $A \cup B \cup C = S$ અને $A \cap B = B \cap C = C \cap A = \emptyset $ . The number of ways to partition $S$ is

If ${ }^{1} \mathrm{P}_{1}+2 \cdot{ }^{2} \mathrm{P}_{2}+3 \cdot{ }^{3} \mathrm{P}_{3}+\ldots+15 \cdot{ }^{15} \mathrm{P}_{15}={ }^{\mathrm{q}} \mathrm{P}_{\mathrm{r}}-\mathrm{s}, 0 \leq \mathrm{s} \leq 1$ then ${ }^{\mathrm{q}+\mathrm{s}} \mathrm{C}_{\mathrm{r}-\mathrm{s}}$ is equal to .... .

The number of ways in which $3$ children can distribute $10$ tickets out of $15$ consecutively numbered tickets themselves such that they get consecutive blocks of $5, 3$ and $2$ tickets is