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Q. Find $r$, if $5 ^{4}P_{r} = 6 ^{5}Pr-1$ .

Permutations and Combinations

Solution:

We have $5 ^{4}P_{r} = 6 ^{5}P_{r-1} $
$\Rightarrow 5\times \frac{4!}{\left(4-r\right)!} = 6\times\frac{5!}{\left(5-r+1\right)!} $
$ \Rightarrow r^{2}-11r + 24 = 0$
$\Rightarrow r =8$ or $r=3$.