Prove that:
(i) $\frac{n !}{(n-r) !}=n(n-1)(n-2) \ldots(n-(r-1))$
(ii) $\frac{n !}{(n-r) ! r !}+\frac{n !}{(n-r+1) !(r-1) !}=\frac{(n+1) !}{r !(n-r+1) !}$
(i) $\mathrm{LHS}=\frac{n !}{(n-r) !}$
$=\frac{n(n-1)(n-2)(n-3)(n-4) \ldots(n-r+1)[(n-r) !]}{(n-r) !}$
$=n(n-1)(n-2)(n-3)(n-4) \ldots(n-r+1)$
$=n(n-1)(n-2)(n-3)(n-4) \ldots[n-(r-1)]=$ RHS
(ii) $\mathrm{LHS}=\frac{n !}{(n-\mathrm{r}) ! \mathrm{r} !}+\frac{n !}{(n-\mathrm{r}+1) !}$
$=\frac{n !}{(n-r) ! r !}+\frac{n !}{(n-r+1)[(n-r) !]}$
$=\frac{n !(n-r+1)+n ! r !}{r !(n-r+1)[(n-r) !]}$
$=\frac{n !(n+1)-n ! r !+n ! r !}{r !(n-r+1)(n-r) !}$
$=\frac{n !(n+1)}{r !(n-r+1)(\mathrm{n}-\mathrm{r}) !}$
$=\frac{(n+1 !)}{r !(n-r+1) !}=\mathrm{RHS}$
Hence proved.
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