Question

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Answer 1

★ Solution ★

★To prove:-

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

★Proof:-

LHS:-

$:\implies \sf \frac{n!}{r!(n-r)!} +\frac{n!}{(r-1)!(n-r+1)!} \\\\We\ know,\\\\\leadsto [r!=r(r-1)] \\ And,\\\\\leadsto (n-r+1)!=(n-r+1)(n-r+1-1)\\\\=(n-r-1)(n-r)!\\\\Now,\\\\:\implies \sf \frac{n!}{r(r-1)!(n-r)!} +\frac{n!}{(r-1)!(n-r+1)(n-r)!}$

Taking common,

$:\implies \sf \frac{n!}{(r-1)!} [\frac{1}{r} +\frac{1}{n-r+1} ]\\\\:\implies \sf \frac{n!}{(r-1)!(n-r)!} [\frac{n-\cancel{r}+1+\cancel{r}}{r(n-r+1)} ]\\\\:\implies \sf \frac{n!(n+1)}{r(r-1)!(n-r+1)(n-r)!}$

Using (n+1)! = (n+1)n! ,

$\mapsto \underline{\boxed{\sf \frac{(n+1)!}{r!(n-r+1)!} }}$

Hence proved !

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Answer 2

To prove :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Proof :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

LHS :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$:\implies \sf \dfrac{n!}{r!(n-r)!} +\dfrac{n!}{(r-1)!(n-r+1)!} \\\\\\ \sf We\ know,\\\\\\ \sf \implies [r!=r(r-1)] \\\\ \sf And,\\\\\\ \sf \implies (n-r+1)!=(n-r+1)(n-r+1-1)\\\\\\ \sf =(n-r-1)(n-r)!\\\\\\ \sf Now,\\\\\\:\implies \sf \frac{n!}{r(r-1)!(n-r)!} +\frac{n!}{(r-1)!(n-r+1)(n-r)!}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Taking common,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \dfrac{n!}{(r-1)!} [\dfrac{1}{r} +\dfrac{1}{n-r+1} ]\\\\:\implies \sf \dfrac{n!}{(r-1)!(n-r)!} [\dfrac{n-\cancel{r}+1+\cancel{r}}{r(n-r+1)} ]\\\\:\implies \sf \dfrac{n!(n+1)}{r(r-1)!(n-r+1)(n-r)!}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Using (n+1)! = (n+1)n!,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\implies{\sf \dfrac{(n+1)!}{r!(n-r+1)!} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Hence proved..