Prove that

Question:

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) $\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}+\frac{\mathrm{n} !}{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}+1) !}=\frac{(\mathrm{n}+1) !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}+1) !}$

Solution:

(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.

(ii) To Prove $(n-r+1) \cdot \frac{n !}{(n-r+1) !}=\frac{n !}{(n-r) !}$

Formula: $n !=n \times(n-1) !$

L.H.S. $=(n-r+1) \frac{n !}{(n-r+1) !}$

By using above formula,

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

Cancelling (n - r + 1),

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

$=$ R.H.S.

$\therefore \mathrm{LHS}=\mathrm{RHS}$

(iii) To Prove  $\frac{n !}{(r !) \times(n-r) !}+\frac{n !}{(r-1) ! \times(n-r+1) !}=\frac{(n+1) !}{(r !) \times(n-r+1) !}$

Formula : $n !=n \times(n-1) !$

L.H.S. $=\frac{n !}{(r !) \times(n-r) !}+\frac{n !}{(r-1) ! \times(n-r+1) !}$

By using above formula,

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

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

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

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

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

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

$=$ R.H.S.

$\therefore \mathrm{LHS}=\mathrm{RHS}$