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How do you find the value of the permutation P(12,0)?

1 Answer

Jun 4, 2017

$1$ $1$

Explanation:

$using the definition of^{n} P_{r} = \frac{n!}{(n - r)!}$ $"using the definition of " ""^nP_r=(n!)/((n-r)!)$

$\Rightarrow^{12} P_{0} = \frac{12!}{(12 - 0)!} = \frac{12!}{12!} = 1$ $rArr""^(12)P_0=(12!)/((12-0)!)=(12!)/(12!)=1$

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