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Question Number 25314 by ibraheem160 last updated on 08/Dec/17
prove that 0!=1
$${prove}\:{that}\:\mathrm{0}!=\mathrm{1} \\ $$
Commented by prakash jain last updated on 08/Dec/17
please see question 25237.
$$\mathrm{please}\:\mathrm{see}\:\mathrm{question}\:\mathrm{25237}. \\ $$
Answered by $@ty@m last updated on 09/Dec/17
Wehave  ^1 P_1 =((1!)/((1−1)!))  (by def.)  ^1 P_1 =(1/(0!))  −−−(1)  ^1 P_1 = no. of arrangement of 1   thing out of 1  ⇒^1 P_1 =1 −−−(2)  From (1) and (2)  0!=1
$${Wehave} \\ $$$$\:^{\mathrm{1}} {P}_{\mathrm{1}} =\frac{\mathrm{1}!}{\left(\mathrm{1}−\mathrm{1}\right)!}\:\:\left({by}\:{def}.\right) \\ $$$$\:^{\mathrm{1}} {P}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{0}!}\:\:−−−\left(\mathrm{1}\right) \\ $$$$\:^{\mathrm{1}} {P}_{\mathrm{1}} =\:{no}.\:{of}\:{arrangement}\:{of}\:\mathrm{1}\: \\ $$$${thing}\:{out}\:{of}\:\mathrm{1} \\ $$$$\Rightarrow\:^{\mathrm{1}} {P}_{\mathrm{1}} =\mathrm{1}\:−−−\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{0}!=\mathrm{1} \\ $$

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