prove that: n! (n-r)!r! + n! (n-r+1)!(r-1)! = (n+1)! r(n-r+1)! - Vidyadip

Question

prove that:
$\frac{\text{n}!}{\text{(n-r)!r!}}+\frac{\text{n!}}{\text{(n-r+1)!}\text{(r-1)!}}= \frac{\text{(n+1)!}}{\text{r(n-r+1)!}}$

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Answer

We have,
$\text{L.H.S.}=\frac{\text{n}!}{\text{(n-r)!r!}}+\frac{\text{n!}}{\text{(n-r+1)!}\text{(r-1)!}}$
$=\frac{\text{n!}}{(\text{n-r})!\text{r}\times\big[\text{(r-1)}!\big]}+\frac{\text{n!}}{\text{(n-r+1)}\big[(\text{n-r})!\big](\text{r-1})!}$
$=\frac{\text{n!}}{\text{(n - r)! }\times{\text{(r-1)!}}}\bigg[\frac{1}{\text{r}}+\frac{1}{\text{n-r+1}}\bigg]$
$=\frac{\text{n!}}{\text{(n - r)! }\times{\text{(r-1)!}}}\bigg[\frac{\text{n-r+1+r}}{\text{r(n-r+1)}}\bigg]$
$=\frac{\text{n!}}{\text{(n - r)! }\times{\text{(r-1)!}}}\bigg[\frac{\text{n+1}}{\text{r(n-r+1)}}\bigg]$
$=\frac{\text{(n+1)}\times\text{n!}}{\text{(n-r+1)!}\times\text{(n-r)}! \times\text{r}\times(\text{r-1})!}$
$=\frac{(\text{n+1}!)}{\text{(n-r+1)!}\times\text{r}!}$
$=\frac{\text{(n+1)!}}{\text{r!}\text{(n-r+1)}!}$
$\text{R.H.S}$
$\therefore$ L.H.S.= R.H.S.

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