Question
Find $r$ if ${ }^{12} \mathrm{P}_{\mathrm{r}-2}:{ }^{11} \mathrm{P}_{\mathrm{r}-1}=3: 14$

Answer

$
\begin{aligned}
& { }^{12} \mathrm{P}_{\mathrm{r}-2}:{ }^{11} \mathrm{P}_{\mathrm{r}-1}=3: 14 \\
& \therefore \quad \frac{12 !}{(12-r+2) !} \div \frac{11 !}{(11-r+1) !}=\frac{3}{14} \\
& \therefore \quad \frac{12 !}{(14-r) !} \times \frac{(12-r) !}{11 !}=\frac{3}{14} \\
& \therefore \quad \frac{12 \times 11 !}{(14-r)(13-r)(12-r) !} \times \frac{(12-r) !}{11 !}=\frac{3}{14} \\
& \therefore \quad \frac{12}{(14-r)(13-r)}=\frac{3}{14} \text { MaharashtraBoardSolutions.in } \\
& \therefore(14-r)(13-r)=8 \times 7 \\
&
\end{aligned}
$
Comparing on both sides, we get
$
\begin{aligned}
& 14-r=8 \\
& \therefore r=6
\end{aligned}
$

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