Question

Show that

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

Answer 1

★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:-

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

★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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