Haryana Police Constable Model Paper 2

Given:
${}_{}^{\mathrm{n}}{\mathrm{P}}_{\mathrm{r}}^{}$ = 990 and ${}_{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}^{}$ = 165
Formula used:
${}_{}^{\mathrm{n}}{\mathrm{P}}_{\mathrm{r}}^{}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}$
${}_{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}^{}$ = $\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!\times \mathrm{r}!}$
Calculation:
${}_{}^{\mathrm{n}}{\mathrm{P}}_{\mathrm{r}}^{}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}$ ….(i)
${}_{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}^{}$ = $\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!\times \mathrm{r}!}$ ….(ii)
Divide equation (i) and (ii)
⇒ $\frac{{}^{{\mathrm{n}}_{}}{\mathrm{P}}_{\mathrm{r}}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}$ = $\frac{\frac{\mathrm{n}}{\left(\mathrm{n}-\mathrm{r}\right)!}}{\frac{\mathrm{n}!}{\left[\frac{\mathrm{n}}{\left(\mathrm{n}-\mathrm{r}\right)!\times \mathrm{r}!}\right]}}$
$\Rightarrow \frac{{}^{{\mathrm{n}}_{}}{\mathrm{P}}_{\mathrm{r}}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}$ = r!

⇒ $\frac{990}{165}$ = r!
⇒ r! = 6
⇒ r! = 3 × 2 × 1
⇒ r = 3
The value of r putting in ${}_{}^{\mathrm{n}}{\mathrm{P}}_{\mathrm{r}}^{}$=990
⇒ ${}_{}^{\mathrm{n}}{\mathrm{P}}_3^{}$ = 990
⇒ n × (n – 1) × (n – 2) = 11 × 10 × 9
From this we get
n = 11
∴Values of n and r are respectively 11 and 3