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Solution:

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,

${ }_{{ }^{\mathrm{n}} \operatorname{Pr}}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}$

${ }^{\mathrm{n}} \mathrm{P}_{4}:{ }^{\mathrm{n}} \mathrm{P}_{5}=1: 2$

$\frac{n !}{(n-4) !}: \frac{n !}{(n-5) !}=\frac{1}{2}$

$\frac{n !}{(n-4)(n-5) !}: \frac{n !}{(n-5) !}=\frac{1}{2}$

$\frac{n !}{(n-4)(n-5) !} \times \frac{(n-5) !}{n !}=\frac{1}{2}$

$\frac{1}{(n-4)}=\frac{1}{2}$

$n-4=2$

$n=6$

Hence, value of n is 6.

(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,

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

${ }^{n-1} P_{3}:{ }^{n+1} P_{3},=5: 12$

$\frac{(n-1) !}{(n-4) !}: \frac{(n+1) !}{(n-2) !}=\frac{5}{12}$

$\frac{(n-1) !}{(n-4) !}: \frac{(n+1) n(n-1) !}{(n-2)(n-3)(n-4) !}=\frac{5}{12}$

$\frac{(n-1) !}{(n-4) !} \times \frac{(n-2)(n-3)(n-4) !}{(n+1) n(n-1) !}=\frac{5}{12}$

$\frac{(n-2)(n-3)}{(n+1) n}=\frac{5}{12}$

$\frac{n^{2}-5 n+6}{n^{2}+n}=\frac{5}{12}$

$12^{n^{2}}-60 n+72=5 n^{2}+5 n$

$7 n^{2}-65 n+72=0$

$n=8,2.25$

Since n cannot be 2.25 as it creates a negative factorial in denominator. Therefore, n = 2.25 is not possible.

Hence, value of $n$ is 8 .