If ${ }^{15} \operatorname{Pr}-1:{ }^{16} \operatorname{Pr}-2,=3: 4$, find $\mathrm{r}$.
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,
${ }_{n p_{r}}=\frac{n !}{(n-r) !}$
${ }^{15} P_{r-1}:{ }^{16} P_{r-2},=3: 4$
$\frac{15 !}{(16-r) !}: \frac{16 !}{(18-r) !}=\frac{3}{4}$
$\frac{15 !}{(16-\mathrm{r}) !}: \frac{16 \times 15 !}{(18-\mathrm{r})(17-\mathrm{r})(16-\mathrm{r}) !}=\frac{3}{4}$
$\frac{15 !}{(16-\mathrm{r}) !} \times \frac{(18-\mathrm{r})(17-\mathrm{r})(16-\mathrm{r}) !}{16 \times 15 !}=\frac{3}{4}$
$\frac{(18-\mathrm{r})(17-\mathrm{r})}{4}=3$
$\frac{(18-r)(17-r)}{16}=\frac{3}{4}$
$r^{2}-35 r+306=12$
$r^{2}-35 r+294=0$
$(r-21)(r-14)=0$
$r=21,14$
Since $r$ cannot be 21 as it creates a negative factorial in denominator. Therefore, $\mathrm{r}=14$ is not possible.
Hence, value of r is 14