a) $n= 2 \left(1+1\right),\;n=3\left(1+2, 2+1\right),\\n=4\left(1+3,3+1,2+2\right),\;n=5\left(1+4,4+1,2+3,3+2\right)$
so $x_1+x_2=n\;\text{and}\;x_1,x_2>{0}$ (no.of integral sol)
This is same as number of ways of putting $\left(n-2\right)$ (as we can't have $0$ for either $x_1$ or $x_2$) identical balls into two distinct bins, which is obtained by putting a divider across $\left(n-2\right)$ balls and taking all possible permutations with $\left(n-2\right)$ being identical. i.e., $\frac{(n-2 + 1)!}{(n-2)!} = (n-1).$
We can also use the following formula ,
$^{(n-2+2-1)}C_{(2-1)}=^{n-1}C_1.$
b) $n=3\left(1+1+1\right),\;n=4\left(1+1+2,1+2+1,2+1+1\right),\\ n=5\left(1+1+3,1+3+1,3+1+1,2+2+1,2+1+2,1+2+2\right) $
so $x_1+x_2+x_3=n\;\text{and}\;x_1,x_2,x_3>0$ (no.of integral sol)
Here, we can permute $\left(n-3\right)$ items with $2$ dividers which will give $\frac{(n-3 + 2)!}{(n-3)!2!}$
$\begin{align}&=\frac{\left(n-1\right)!}{\left(n-1-2\right)!2!}\\\\&=\;^{n-1}C_2\end{align}$
c) $^{(n-k+k-1)}C_{k-1}=^{n-1}C_{k-1}.$