Prove that
(i) $\frac{n !}{r !}=n(n-1)(n-2) \ldots(r+1)$
(ii) $(n-r+1)$ $\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}$
(iii) $\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}+\frac{\mathrm{n} !}{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}+1) !}=\frac{(\mathrm{n}+1) !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}+1) !}$
(i) To Prove $\frac{n !}{r !}=n(n-1)(n-2) \ldots(r+1)$
Formula: $n !=n \times(n-1) !$
L.H.S. $=\frac{n !}{r !}$
Writing (n!) in terms of (r!) by using above formula
$=\frac{n(n-1)(n-2) \ldots \ldots(r+1)(r !)}{r !}$
Cancelling (r!),
$=n(n-1)(n-2) \ldots(r+1)$
$=$ R.H.S.
$\therefore \mathrm{LHS}=\mathrm{RHS}$
Note : In permutation and combination $\mathrm{r}$ is always less than $\mathrm{n}$, so we can write $\mathrm{n} !$ in terms of r! by using given formula.
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