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How-many-3-digit-numbers-divisible-by-5-can-be-made-using-the-numbers-7-8-9-2-1-0-without-repetition-




Question Number 125182 by Mammadli last updated on 08/Dec/20
How many 3−digit numbers divisible by 5 can be made using the numbers 7,8,9,2,1,0 without repetition?
$$\boldsymbol{{How}}\:\boldsymbol{{many}}\:\mathrm{3}−\boldsymbol{{digit}}\:\boldsymbol{{numbers}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{5}\:\boldsymbol{{can}}\:\boldsymbol{{be}}\:\boldsymbol{{made}}\:\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{numbers}}\:\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{2},\mathrm{1},\mathrm{0}\:\boldsymbol{{without}}\:\boldsymbol{{repetition}}? \\ $$
Answered by bemath last updated on 08/Dec/20
AB0 ⇒ P_2 ^( 5)  = ((5!)/(3!)) = 20.
$${AB}\mathrm{0}\:\Rightarrow\:{P}_{\mathrm{2}} ^{\:\mathrm{5}} \:=\:\frac{\mathrm{5}!}{\mathrm{3}!}\:=\:\mathrm{20}. \\ $$

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