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Question Number 55119 by pieroo last updated on 17/Feb/19

How many numbers, divisible by 5, can be made with the digits 2,3,4 and 5 where no digit is being used more than once in each number?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers},\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5},\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{made}\:\mathrm{with}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{2},\mathrm{3},\mathrm{4}\:\mathrm{and}\:\mathrm{5}\:\mathrm{where} \\ $$$$\mathrm{no}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{being}\:\mathrm{used}\:\mathrm{more}\:\mathrm{than}\:\mathrm{once} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{number}? \\ $$

Commented by Tawa1 last updated on 17/Feb/19

For a number to be divisible by 5. The number must end with 5. ∴ for all the four digit number (2, 3, 4, 5). 3 × 2 × 1 (fixed 5) = 6 ways − − − − ∴ for three digit number (2, 3, 4, 5). 3 × 2 (fixed 5) = 6 ways − − − ∴ for two digit number (2, 3, 4, 5). 3 (fixed 5) = 3 ways − − ∴ for one digit number (2, 3, 4, 5). (only 5) = 1 way − Therefore, Number of numbers divisible by 5 = 6 + 6 + 3 + 1 Therefore, Number of numbers divisibke by 5 = 16 numbers

$$\mathrm{For}\:\mathrm{a}\:\mathrm{number}\:\mathrm{to}\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\:\mathrm{5}.\:\:\mathrm{The}\:\mathrm{number}\:\mathrm{must}\:\mathrm{end}\:\mathrm{with}\:\mathrm{5}. \\ $$$$\therefore\:\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{the}\:\mathrm{four}\:\mathrm{digit}\:\mathrm{number}\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:×\:\mathrm{2}\:×\:\mathrm{1}\:\left(\mathrm{fixed}\:\mathrm{5}\right)\:\:\:=\:\:\mathrm{6}\:\mathrm{ways} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:−\:\:−\:\:− \\ $$$$\therefore\:\:\:\mathrm{for}\:\mathrm{three}\:\mathrm{digit}\:\mathrm{number}\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:×\:\mathrm{2}\:\left(\mathrm{fixed}\:\mathrm{5}\right)\:\:\:=\:\:\mathrm{6}\:\mathrm{ways} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:−\:\:−\: \\ $$$$\therefore\:\:\:\mathrm{for}\:\mathrm{two}\:\mathrm{digit}\:\mathrm{number}\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\left(\mathrm{fixed}\:\mathrm{5}\right)\:\:\:=\:\:\mathrm{3}\:\mathrm{ways} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:−\:\:\: \\ $$$$\therefore\:\:\:\mathrm{for}\:\mathrm{one}\:\mathrm{digit}\:\mathrm{number}\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{only}\:\mathrm{5}\right)\:\:\:=\:\:\mathrm{1}\:\mathrm{way} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\: \\ $$$$\mathrm{Therefore},\:\:\:\:\:\mathrm{Number}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5}\:\:\:=\:\:\mathrm{6}\:+\:\mathrm{6}\:+\:\mathrm{3}\:+\:\mathrm{1} \\ $$$$\mathrm{Therefore},\:\:\:\:\:\mathrm{Number}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{divisibke}\:\mathrm{by}\:\mathrm{5}\:\:\:=\:\:\mathrm{16}\:\:\mathrm{numbers} \\ $$

Answered by math1967 last updated on 17/Feb/19

3!=6 (4 digit number) P_2 ^3 =6(3digit numbers) P_1 ^3 =3 (2digit numbers) 1digit number =1 total=6+6+3+1=16

$$\mathrm{3}!=\mathrm{6}\:\left(\mathrm{4}\:{digit}\:{number}\right) \\ $$$$\overset{\mathrm{3}} {{P}}_{\mathrm{2}} =\mathrm{6}\left(\mathrm{3}{digit}\:{numbers}\right) \\ $$$$\:\overset{\mathrm{3}} {{P}}_{\mathrm{1}} =\mathrm{3}\:\left(\mathrm{2}{digit}\:{numbers}\right) \\ $$$$\:\mathrm{1}{digit}\:{number}\:=\mathrm{1} \\ $$$${total}=\mathrm{6}+\mathrm{6}+\mathrm{3}+\mathrm{1}=\mathrm{16} \\ $$

Commented by pieroo last updated on 18/Feb/19

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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