# Find r if (i) (ii) .

Question:

Find $r$ if

(i) ${ }^{5} \mathrm{P}_{r}=2^{6} \mathrm{P}_{r-1}$

(ii) ${ }^{3} \mathrm{P}_{n}={ }^{6} \mathrm{P}_{r-1}$

Solution:

(i)

${ }^{5} \mathrm{P}_{r}=2^{6} \mathrm{P}_{r-1}$

$\Rightarrow \frac{5 !}{(5-r) !}=2 \times \frac{6 !}{(6-r+1) !}$

$\Rightarrow \frac{5 !}{(5-r) !}=\frac{2 \times 6 !}{(7-r) !}$

$\Rightarrow \frac{5 !}{(5-r) !}=\frac{2 \times 6 \times 5 !}{(7-r)(6-r)(5-r) !}$

$\Rightarrow 1=\frac{2 \times 6}{(7-r)(6-r)}$

$\Rightarrow(7-r)(6-r)=12$

$\Rightarrow 42-6 r-7 r+r^{2}=12$

$\Rightarrow r^{2}-13 r+30=0$

$\Rightarrow r^{2}-3 r-10 r+30=0$

$\Rightarrow r(r-3)-10(r-3)=0$

$\Rightarrow(r-3)(r-10)=0$

$\Rightarrow(r-3)=0$ or $(r-10)=0$

$\Rightarrow r=3$ or $r=10$

it is known that, ${ }^{n} \mathrm{P}_{r}=\frac{n !}{(n-r) !}$, where $0 \leq r \leq n$

$\therefore 0 \leq r \leq 5$

Hence, $r \neq 10$

$\therefore r=3$

(ii)

${ }^{5} \mathrm{P}_{r}={ }^{6} \mathrm{P}_{r-1}$

$\Rightarrow \frac{5 !}{(5-r) !}=\frac{6 !}{(6-r+1) !}$

$\Rightarrow \frac{5 !}{(5-r) !}=\frac{6 \times 5 !}{(7-r) !}$

$\Rightarrow \frac{1}{(5-r) !}=\frac{6}{(7-r)(6-r)(5-r) !}$

$\Rightarrow 1=\frac{6}{(7-r)(6-r)}$

$\Rightarrow(7-r)(6-r)=6$

$\Rightarrow 42-7 r-6 r+r^{2}-6=0$

$\Rightarrow r^{2}-13 r+36=0$

$\Rightarrow r^{2}-4 r-9 r+36=0$

$\Rightarrow r(r-4)-9(r-4)=0$

$\Rightarrow(r-4)(r-9)=0$

$\Rightarrow(r-4)=0$ or $(r-9)=0$

$\Rightarrow r=4$ or $r=9$

It is known that, ${ }^{n} \mathrm{P}_{r}=\frac{n !}{(n-r) !}$, where $0 \leq r \leq n$

$\therefore 0 \leq r \leq 5$

Hence, $r \neq 9$

$\therefore r=4$