$\left(A\right)$  The number of ways of not selecting $\left(n-r\right)$ things from $n$ different things

$={}^{n}C_r={}^{n}C_{n-r}$

$\left(B\right)$ $\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{{\displaystyle {}^{n}C_{r-1}}}{{\displaystyle {}^{n}C_{n-r}}}$

$\Rightarrow \frac{{}^{n}C_{r-1}}{{}^{n}C_{n-r}}=\frac{n!}{\left(r-1\right)!\times \left(n-r+1\right)!}\times \frac{r!\times \left(n-r\right)!}{n!}$

$\Rightarrow \frac{{}^{n}C_{r-1}}{{}^{n}C_{n-r}}=\frac{r}{\left(n-r+1\right)}$

$\Rightarrow \left(n-r+1\right){}^{n}C_{r-1}=r\left({}^{n}C_{n-r}\right)$

$\Rightarrow \left(n-r+1\right){}^{n}C_{r-1}=r\left({}^{n}C_r\right)$

$\left(C\right)$ Number of ways of selecting atleast $\left(n-r\right)$ tnings from $n$ different things

$=\left({}^{n}C_0+{}^{n}C_1+{}^{n}C_2+....+{}^{n}C_{n-r}+{}^{n}C_{n-r+1}+...+{}^{n}C_n\right)-\left({}^{n}C_0+{}^{n}C_1+{}^{n}C_2+....+{}^{n}C_{n-r-1}\right)$

$={}^{n}C_{n-r}+{}^{n}C_{n-r+1}+{}^{n}C_{n-r+2}+......+{}^{n}C_{n-1}+{}^{n}C_n$

$={}^{n}C_0+{}^{n}C_1+.......+{}^{n}C_{r-2}{}^{n}C_{r-2}+{}^{n}C_{r-1}+{}^{n}C_r$

$\left[\because {}^{n}C_r={}^{n}C_{n-r}\right]$

$\left(D\right)$

$\left(n-r\right)\left({}^{n-1}C_{r-1}+{}^{n-1}C_r\right)$

$=\left(n-r\right)\left({}^{n}C_r\right)$

$=\left(n-r\right)\left(\frac{n!}{r!\times \left(n-r\right)!}\right)$

$=\left(\frac{n!}{r!\times \left(n-r-1\right)!}\right)$

$=\left(r+1\right)\left(\frac{n!}{\left(r+1\right)!\times \left(n-r-1\right)!}\right)$

$=\left(r+1\right){}^{n}C_{r+1}$