Q58 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

If ${ }^n P_r={ }^{(n-1)} P_r+x^{(n-1)} P_{(r+1)}, \forall n, r \in N$ and $r \leq n$, then $x=$

  • A
    $(n+1)$
  • B
    $(r+1)$
  • C
    $n$
  • Correct answer
    D
    $r$

Solution

$\begin{aligned}{ }^n P_r- & { }^{n-1} P_r=\frac{n !}{(n-r) !}-\frac{(n-1) !}{(n-r-1) !} \\ & =\frac{(n-1) !}{(n-r-1) !}\left(\frac{n}{n-r}-1\right)=\frac{(n-1) !}{(n-r) !} \cdot r \\ & =\frac{(n-1) !}{((n-1)-(r-1) !} \cdot r=r \cdot{ }^{n-1} P_{r-1} \\ \therefore \quad{ }^n P_r & ={ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1} \\ \therefore \quad x & =r\end{aligned}$

Hence, the correct answer is option: 'D'.

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