Q64 / 160 · AP EAMCET 2011 · Mathematics

${ }^{15} P_8=A+8 \cdot{ }^{14} P_7 \Rightarrow A=$

  • A
    ${ }^{14} P_6$
  • Correct answer
    B
    ${ }^{14} P_8$
  • C
    ${ }^{15} P_7$
  • D
    ${ }^{16} P_9$

Solution

$\begin{aligned} & \text { Given, }{ }^{15} P_8=A+8 \cdot{ }^{14} P_7 \\ & \Rightarrow \quad \frac{15 !}{7 !}=A+8 \cdot \frac{14 !}{7 !} \\ & \Rightarrow \quad A=\frac{14 !}{7 !}(15-8)=\frac{14 !}{7 !}(7) \\ & =\frac{14 !}{6 !}={ }^{14} P_8 \\ & \end{aligned}$

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

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