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6.Permutation and Combination
easy
यदि ${ }^{ n } P _{ r }={ }^{ n } P _{ r +1}$ तथा ${ }^{ n } C _{ r }={ }^{ n } C _{ I -1}$ है, तो $r$ बराबर है
A
$3$
B
$1$
C
$4$
D
$2$
(JEE MAIN-2021)
Solution
${ }^{n} \mathrm{p}_{r}={ }^{n} p_{r+1} \Rightarrow \frac{n !}{(n-r) !}=\frac{n !}{(n-r-1) !}$
$\Rightarrow(n-r)=1$
${ }^{n} \mathrm{C}_{\mathrm{r}}={ }^{n} C_{\mathrm{r}-1}$
$\Rightarrow \frac{n !}{r !(n-r) !}=\frac{n !}{(r-1) !(n-r+1) !}$
$\Rightarrow \frac{1}{r(n-r) !}=\frac{1}{(n-r+1)(n-r) !}$
$\Rightarrow \mathrm{n}-\mathrm{r}+1=\mathrm{r}$
$\Rightarrow \mathrm{n}+1=2 \mathrm{r}$
$(1)$ $\Rightarrow 2 r-1-r=1 \Rightarrow r=2$
Standard 11
Mathematics
