# Solve this

Question:

If ${ }^{15} \operatorname{Pr}-1:{ }^{16} \operatorname{Pr}-2,=3: 4$, find $\mathrm{r}$.

Solution:

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

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