Question Number 25314 by ibraheem160 last updated on 08/Dec/17
$${prove}\:{that}\:\mathrm{0}!=\mathrm{1} \\ $$
Commented by prakash jain last updated on 08/Dec/17
$$\mathrm{please}\:\mathrm{see}\:\mathrm{question}\:\mathrm{25237}. \\ $$
Answered by $@ty@m last updated on 09/Dec/17
$${Wehave} \\ $$$$\:^{\mathrm{1}} {P}_{\mathrm{1}} =\frac{\mathrm{1}!}{\left(\mathrm{1}−\mathrm{1}\right)!}\:\:\left({by}\:{def}.\right) \\ $$$$\:^{\mathrm{1}} {P}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{0}!}\:\:−−−\left(\mathrm{1}\right) \\ $$$$\:^{\mathrm{1}} {P}_{\mathrm{1}} =\:{no}.\:{of}\:{arrangement}\:{of}\:\mathrm{1}\: \\ $$$${thing}\:{out}\:{of}\:\mathrm{1} \\ $$$$\Rightarrow\:^{\mathrm{1}} {P}_{\mathrm{1}} =\mathrm{1}\:−−−\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{0}!=\mathrm{1} \\ $$