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How-many-three-digit-numbers-can-be-formed-using-the-digits-2-3-5-7-8-if-the-number-is-odd-and-no-digit-is-repeted-




Question Number 153714 by pete last updated on 09/Sep/21
How many three−digit numbers can be  formed using the digits 2,3,5,7,8 if   the number is odd and no digit is repeted?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{three}−\mathrm{digit}\:\mathrm{numbers}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{formed}\:\mathrm{using}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{8}\:\mathrm{if}\: \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{no}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{repeted}? \\ $$
Answered by EDWIN88 last updated on 09/Sep/21
= P_1 ^( 3) ×P_2 ^( 4)  =((3!)/(2!))×((4!)/(2!)) = 6×6=36
$$=\:{P}_{\mathrm{1}} ^{\:\mathrm{3}} ×{P}_{\mathrm{2}} ^{\:\mathrm{4}} \:=\frac{\mathrm{3}!}{\mathrm{2}!}×\frac{\mathrm{4}!}{\mathrm{2}!}\:=\:\mathrm{6}×\mathrm{6}=\mathrm{36} \\ $$
Answered by puissant last updated on 09/Sep/21
A_3 ^1 ×A_4 ^2 =((3!)/((3−1)!))×((4!)/((4−2)!))  =3×4×3  =36..
$${A}_{\mathrm{3}} ^{\mathrm{1}} ×{A}_{\mathrm{4}} ^{\mathrm{2}} =\frac{\mathrm{3}!}{\left(\mathrm{3}−\mathrm{1}\right)!}×\frac{\mathrm{4}!}{\left(\mathrm{4}−\mathrm{2}\right)!} \\ $$$$=\mathrm{3}×\mathrm{4}×\mathrm{3} \\ $$$$=\mathrm{36}.. \\ $$

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