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Question Number 124372 by mr W last updated on 02/Dec/20
How many six-digit numbers contain  exactly six different digits?
$${How}\:{many}\:{six}-{digit}\:{numbers}\:{contain} \\ $$$${exactly}\:{six}\:{different}\:{digits}? \\ $$
Answered by bemath last updated on 02/Dec/20
= 9×P_5 ^( 9)  = 9×((9!)/(4!))   = 9×9×8×7×6×5=136080
$$=\:\mathrm{9}×{P}_{\mathrm{5}} ^{\:\mathrm{9}} \:=\:\mathrm{9}×\frac{\mathrm{9}!}{\mathrm{4}!} \\ $$$$\:=\:\mathrm{9}×\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}=\mathrm{136080} \\ $$
Commented by mr W last updated on 03/Dec/20
thanks!
$${thanks}! \\ $$

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