यदि { }^{ n } P _{ r }={ }^{ n } P _{ r +1} त | 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

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

यदि ${ }^{ n } P _{ r }={ }^{ n } P _{ r +1}$ तथा ${ }^{ n } C _{ r }={ }^{ n } C _{ I -1}$ है, तो $r$ बराबर है

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