Question

If ${}^6{\mathrm{P}}_{\mathrm{r}}=360$ and if ${}^6\mathrm{C}_{\mathrm{r}}=15$, find $\mathrm{r}$?

Answer 1

Find the value of $\mathrm{r}$.

We know that, ${}^{\mathrm{n}}\mathrm{P}_{\mathrm{r}}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}\times \mathrm{r}!$

Put, $n=6$, we get,

${}^6\mathrm{P}_{\mathrm{r}}={}^6\mathrm{C}_{\mathrm{r}}\times \mathrm{r}!$

Put ${}^6{\mathrm{P}}_{\mathrm{r}}=360$ and ${}^6\mathrm{C}_{\mathrm{r}}=15$, we get,

$360=15\times \mathrm{r}!$

$\Rightarrow \mathrm{r}!=\frac{360}{15}$

$\Rightarrow \mathrm{r}!=24$

We know that $4!=4\times 3\times 2\times 1=24$

$\Rightarrow \mathrm{r}!=4!$

$\Rightarrow \mathrm{r}=4$

Hence, the value of $\mathrm{r}$ is $4$.