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Question Number 1740 by tabrez8590@gmail last updated on 11/Sep/15
how  0!=1
$${how}\:\:\mathrm{0}!=\mathrm{1} \\ $$
Commented by 123456 last updated on 11/Sep/15
1!=1∙0!  so for it work to 1 them 0!=1
$$\mathrm{1}!=\mathrm{1}\centerdot\mathrm{0}! \\ $$$$\mathrm{so}\:\mathrm{for}\:\mathrm{it}\:\mathrm{work}\:\mathrm{to}\:\mathrm{1}\:\mathrm{them}\:\mathrm{0}!=\mathrm{1} \\ $$
Answered by Rasheed Ahmad last updated on 12/Sep/15
Permutation of r elements from  a set of n elements can be proved  equal to :   P(n,r)=n(n−1)(n−2)....(n−r+1).      For r<n this is equivalent to:    P(n,r)=((n!)/((n−r)!))  As said above for r<n the above  two formulae are consistent.  But at r=n the above formulae  give the following two results  respectively:      P(n,n)=n!       P(n,n)=((n!)/((n−n)!))=((n!)/(0!))     To continue consistency at n=r  we are forced to define:      0! = 1  Rasheed Soomro
$${Permutation}\:{of}\:{r}\:{elements}\:{from} \\ $$$${a}\:{set}\:{of}\:{n}\:{elements}\:{can}\:{be}\:{proved} \\ $$$${equal}\:{to}\:: \\ $$$$\:\boldsymbol{\mathrm{P}}\left({n},{r}\right)={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)….\left({n}−{r}+\mathrm{1}\right).\:\:\:\: \\ $$$${For}\:{r}<{n}\:{this}\:{is}\:{equivalent}\:{to}: \\ $$$$\:\:\boldsymbol{\mathrm{P}}\left({n},{r}\right)=\frac{{n}!}{\left({n}−{r}\right)!} \\ $$$${As}\:{said}\:{above}\:{for}\:{r}<{n}\:{the}\:{above} \\ $$$${two}\:{formulae}\:{are}\:{consistent}. \\ $$$${But}\:{at}\:{r}={n}\:{the}\:{above}\:{formulae} \\ $$$${give}\:{the}\:{following}\:{two}\:{results} \\ $$$${respectively}: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{P}}\left({n},{n}\right)={n}! \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{P}}\left({n},{n}\right)=\frac{{n}!}{\left({n}−{n}\right)!}=\frac{{n}!}{\mathrm{0}!} \\ $$$$\:\:\:{To}\:{continue}\:{consistency}\:{at}\:{n}={r} \\ $$$${we}\:{are}\:{forced}\:{to}\:{define}: \\ $$$$\:\:\:\:\mathrm{0}!\:=\:\mathrm{1} \\ $$$$\mathrm{Rasheed}\:\mathrm{Soomro} \\ $$

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