We have ^nP_r=n !/(n r) ! ...(i) ⇒ ^nP_n=n !/0 ! [putting r = n in (i) ] ⇒ n!=n !/0 ! ...

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Inequality

Prove that 0 ...

Question

Prove that $0! = 1.$

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Solution

We have $^{n} P_{r} = \frac{n!}{(n - r)!} . . . (i)$

$\Rightarrow^{n} P_{n} = \frac{n!}{0!}$ [putting r = n in (i) ]

$\Rightarrow n! = \frac{n!}{0!}$ $[∵^{n} P_{n} = n!]$

$\Rightarrow 0! = \frac{n!}{n!} = 1$

Hence, $0! = 1.$

Remark Thus, we have $" P_{r} = \frac{n!}{(n - r)!},$ where $0 \leq r \leq n .$

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