Solve this

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

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,

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

${ }^{n} P_{5}=20 \times{ }^{n} P_{3}$

$\frac{n !}{(n-5) !}=\left(20 \times \frac{n !}{(n-3) !}\right)$

$\frac{1}{(n-5) !}=\left(20 \times \frac{1}{(n-3)(n-4)(n-5) !}\right)$

$1=\left(20 \times \frac{1}{(n-3)(n-4)}\right)$

$20=(n-3)(n-4)$

$n^{2}-7 n+12=20$

$n^{2}-7 n-8=0$

$(n-8)(n+1)=0$

$n=8,-1$

We know, that n cannot be a negative number.

Hence, value of $n$ is 8

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

Formula Used:

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

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

$16 \times{ }^{n} P_{3}=13 \times{ }^{n+1} P_{3}$

$16 \times \frac{n !}{(n-3) !}=\left(13 \times \frac{(n+1) !}{(n-2) !}\right)$

$16 \times \frac{n !}{(n-3) !}=\left(13 \times \frac{(n+1) n !}{(n-2)(n-3) !}\right)$

$16=13 \times \frac{(n+1)}{(n-2)}$

$16 n-32=13 n+13$

$3 n=45$

$n=15$

Hence, value of n is 15.

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

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,

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

${ }^{2 n} P_{3}=100^{\times}{ }^{n} P_{2}$

$\frac{2 n !}{(2 n-3) !}=\left(100 \times \frac{n !}{(n-2) !}\right)$

$\frac{2 n(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !}=\left(100 \times \frac{n(n-1)(n-2) !}{(n-2) !}\right)$

$\frac{2 n(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !}=\left(100 \times \frac{n(n-1)(n-2) !}{(n-2) !}\right)$

$2 n(2 n-1)(2 n-2)=100 \times n(n-1)$

$4 n(2 n-1)(n-1)=100 \times n(n-1)$

$8 n^{2}-4 n-100 n=0$

$8 n^{2}-104 n=0$

$8 n(n-13)=0$

$n=0,13$

We know that n should be greater than zero.

Hence, value of n is 13