Prove that


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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