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$$
\text { If }{ }^{12} P_r={ }^{11} P_6+6 .{ }^{11} P_5 \text {, then find the value of } r \text {. }
$$
\text { If }{ }^{12} P_r={ }^{11} P_6+6 .{ }^{11} P_5 \text {, then find the value of } r \text {. }
$$
Solution:
2491 Upvotes
Verified Answer
Given: ${ }^{12} \mathrm{P}_{\mathrm{r}}={ }^{11} \mathrm{P}_6+6 .{ }^{11} \mathrm{P}_5$
We know that
$$
\begin{aligned}
&{ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1}=r !{ }^n C_r \\
&\therefore \quad{ }^{11} P_6+6 \cdot{ }^{11} P_5=6 !{ }^{12} C_6 \\
&\Rightarrow \quad{ }^{12} P_6=6 !{ }^{12} C_6
\end{aligned}
$$
$\therefore \quad \frac{12 !}{6 !}=6 ! \frac{12 !}{6 ! 6 !}$ which are equal
$$
\therefore \quad r=6
$$
We know that
$$
\begin{aligned}
&{ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1}=r !{ }^n C_r \\
&\therefore \quad{ }^{11} P_6+6 \cdot{ }^{11} P_5=6 !{ }^{12} C_6 \\
&\Rightarrow \quad{ }^{12} P_6=6 !{ }^{12} C_6
\end{aligned}
$$
$\therefore \quad \frac{12 !}{6 !}=6 ! \frac{12 !}{6 ! 6 !}$ which are equal
$$
\therefore \quad r=6
$$
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