# Solve this

Question:

(i) If If ${ }^{5} P_{r}=2 \times{ }^{6} P_{r-1}$, find $r$.

(ii) If ${ }^{20} P_{r}=13 \times{ }^{20} P_{r-1}$, find $r$.

(iii) If ${ }^{11} P_{r}={ }^{12} P_{r-1}$, find $r$.

Solution:

(i) To find: the value of $r$

Formula Used:

Total number of ways in which $n$ objects can be arranged in r places (Such that no object is replaced) is given by,

${ }_{n} P_{r}=\frac{n !}{(n-r) !}$

${ }^{5} P_{r}=2^{\times}{ }^{6} P_{r-1}$

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

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

$1=\left(\frac{12}{(7-\mathrm{r})(6-\mathrm{r})}\right)$

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

r = 10, 3

Hence, value of r is 3, 10

(ii) To find: the value of $r$

Formula Used:

Total number of ways in which $n$ objects can be arranged in r places (Such that no object is replaced) is given by,

${ }_{n P_{r}}=\frac{n !}{(n-r) !}$

${ }^{20} P_{r}=13^{\times}{ }^{20} P_{r-1}$

$\frac{20 !}{(20-\mathrm{r}) !}=\left(13 \times \frac{20 !}{(21-\mathrm{r}) !}\right)$

$\frac{1}{(20-r) !}=\left(13 \times \frac{1}{(21-r)(20-r) !}\right)$

$21-r=13$

$r=8$

Hence, value of r is 8.

(iii) To find: the value of $r$

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

${ }^{n} P_{r}$ $=\frac{n !}{(n-r) !}$

${ }^{11} \mathrm{P}_{\mathrm{r}}={ }^{12} \mathrm{P}_{\mathrm{r}-1}$

$\frac{11 !}{(11-r) !}=\left(\frac{12 !}{(13-r) !}\right)$

$\frac{11 !}{(11-\mathrm{r}) !}=\left(\frac{12 \times 11 !}{(13-\mathrm{r})(12-\mathrm{r})(11-\mathrm{r}) !}\right)$

$1=\frac{12}{(13-\mathrm{r})(12-\mathrm{r})}$

$r^{2}-25 r+144=0$

$(r-16)(r-9)=0$

$r=16,9$

Since r cannot be 16 as it creates a negative factorial in denominator. Therefore, r = 16 is not possible.

Hence, value of r is 9.