TS EAMCET · Maths · Permutation Combination
\({ }^{15} P_8=A+8 \cdot{ }^{14} P_7 \Rightarrow A=\)
- A \({ }^{14} P_6\)
- B \({ }^{14} P_8\)
- C \({ }^{15} P_7\)
- D \({ }^{16} P_9\)
Answer & Solution
Correct Answer
(B) \({ }^{14} P_8\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given, }{ }^{15} P_8=A+8 \cdot{ }^{14} P_7 \\ & \Rightarrow \quad \frac{15 !}{7 !}=A+8 \cdot \frac{14 !}{7 !} \\ & \Rightarrow \quad A=\frac{14 !}{7 !}(15-8)=\frac{14 !}{7 !}(7) \\ & =\frac{14 !}{6 !}={ }^{14} P_8 \\ & \end{aligned}\)
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