(i) If ${ }^{n} P_{4}:{ }^{n} P_{5},=1: 2$, find $n$.
(ii) If ${ }^{n-1} P_{3}:{ }^{n+1} P_{3},=5: 12$, find $n$.
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 .