$(n - r + 1)^{n} P_{r - 1}$ is equal to

^{n - 1} P_{r}

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^{n + 1} P_{r}

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^{n} P_{r}

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^{n} P_{r + 1}

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Solution

The correct option is C $^{n} P_{r}$
Given

$(n - r + 1)^{n} P_{r - 1}$

$= (n - r + 1) \frac{n!}{(n - r + 1)!}$

$= (n - r + 1) \frac{n!}{(n - r + 1) (n - r)!}$

$= \frac{n!}{(n - r)!}$

$=^{n} P_{r}$

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Similar questions

(n + 1)^{n} P_{r} = (n - r + 1)^{(n + 1)} P_{r}

{^{n} P}_{r} = {^{n} P}_{r + 1}

{^{n} C}_{r} = {^{n} C}_{r - 1}

(n, r)

\frac{^{n} P_{r}}{^{n} P_{r - 2}} = (n - r + 2) (n - r + 1)

^{n} P_{r} =^{n} P_{r + 1}

^{n} C_{r} =^{n} C_{r - 1}

^{n} P_{r} =^{n} P_{r}_{+}_{1}

^{n} C_{r} =^{n} C_{r}_{-}_{1}

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