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Zeroes of a Polynomial

If 8Pr=8Pr+...

Question

If $^{8} P_{r} =^{8} P_{r + 1}$ , then find value of $r$ .

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Solution

As we know that,
$^{n} P_{r} = \frac{n!}{(n - r)!}$
Therefore,
$^{8} P_{r} =^{8} P_{r + 1}$
$\frac{8!}{(8 - r)!} = \frac{8!}{(8 - r - 1)!}$
$\frac{1}{(8 - r) (7 - r)!} = \frac{1}{(7 - r)!}$
$\Rightarrow 8 - r = 1$
$\Rightarrow r = 8 - 1 = 7$

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