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Q. ${ }^{15} P_{8}=A+8 \cdot{ }^{14} P_{7} \Rightarrow A=$

EAMCETEAMCET 2011

Solution:

Given, ${ }^{15} P_{8}=A+8 \cdot{ }^{14} P_{7}$
$\Rightarrow \frac{15 !}{7 !}=A+8 \cdot \frac{14 !}{7 !}$
$\Rightarrow A =\frac{14 !}{7 !}(15-8)=\frac{14 !}{7 !}(7) $
$=\frac{14 !}{6 !}={ }^{14} P_{8} $