If $^{5} P_{r} = 2^{6} P_{r - 1}$ find $r$

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Solution

question is : $^{5} P_{r} = 2^{6} P_{r - 1}$
$^{n} P_{x} = \frac{n!}{(n - x)!}$
$\frac{5!}{(5 - r)!} = 2 \times \frac{6!}{(7 - r)!}$
$(7 - r) (6 - r) = 4 \times 3$
$r = 3$

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