Let $\mathrm{f}\left(\mathrm{n}\right)=\sum _{\mathrm{r}=0}^{\mathrm{n}} \sum _{\mathrm{k}=\mathrm{r}}^{\mathrm{n}} \left(\begin{array}{l}\mathrm{k}\\ \mathrm{r}\end{array}\right)$,then  the total number of divisors of $\mathrm{f}\left(9\right)$.

$\begin{array}{l}\sum _{k=r}^n{ }^k{C}_r{=}^r{C}_r{+}^{r+1}{C}_r{+}^{r+2}{C}_r+\cdots {+}^n{C}_r\\ =1{+}^{r+1}{C}_1{+}^{r+2}{C}_2{+}^{r+3}{C}_3+\cdots {+}^n{C}_{n-r}\end{array}$

 

 

  = ${ }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}{=}^{\mathrm{n}+1}{\mathrm{C}}_{r+1}$
$\begin{array}{l}\therefore \mathrm{f}\left(\mathrm{n}\right)=\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}+1}{=}^{\mathrm{n}+1}{\mathrm{C}}_1{+}^{\mathrm{n}+1}{\mathrm{C}}_2{+}^{\mathrm{n}+1}{\mathrm{C}}_3+\dots {+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}\\ {=}^{\mathrm{n}+1}{\mathrm{C}}_0{+}^{\mathrm{n}+1}{\mathrm{C}}_1{+}^{\mathrm{n}+1}{\mathrm{C}}_2+\dots {+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}-1\\ \mathrm{f}\left(\mathrm{n}\right)=\left(2^{\mathrm{n}+1}\right)-1\\ \mathrm{f}\left(9\right)=2^{10}-1=1023=3\cdot 11\cdot 31\end{array}$
Hence, number of divisors are $(1+1)(1+1)(1+1)=8$