If { }^{n} P_{r}={ }^{n} P_{r+1} and { }^{n} | vedclass

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Question

${ }^{n} \mathrm{p}_{r}={ }^{n} p_{r+1} \Rightarrow \frac{n !}{(n-r) !}=\frac{n !}{(n-r-1) !}$

$\Rightarrow(n-r)=1$

${ }^{n} \mathrm{C}_{\mathrm

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Vedclass · Answer

${ }^{n} \mathrm{p}_{r}={ }^{n} p_{r+1} \Rightarrow \frac{n !}{(n-r) !}=\frac{n !}{(n-r-1) !}$

$\Rightarrow(n-r)=1$

${ }^{n} \mathrm{C}_{\mathrm{r}}={ }^{n} C_{\mathrm{r}-1}$

$\Rightarrow \frac{n !}{r !(n-r) !}=\frac{n !}{(r-1) !(n-r+1) !}$

$\Rightarrow \frac{1}{r(n-r) !}=\frac{1}{(n-r+1)(n-r) !}$

$\Rightarrow \mathrm{n}-\mathrm{r}+1=\mathrm{r}$

$\Rightarrow \mathrm{n}+1=2 \mathrm{r}$

$(1)$ $\Rightarrow 2 r-1-r=1 \Rightarrow r=2$

If ${ }^{n} P_{r}={ }^{n} P_{r+1}$ and ${ }^{n} C_{r}={ }^{n} C_{r-1}$, then the value of $r$ is equal to:

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