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

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

Commented by Tawa1 last updated on 17/Feb/19

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

Answered by math1967 last updated on 17/Feb/19

$$3!=6\left(4digitnumber\right)$$$${\stackrel{3}{P}}_2=6\left(3digitnumbers\right)$$$${\stackrel{3}{P}}_1=3\left(2digitnumbers\right)$$$$1digitnumber=1$$$$total=6+6+3+1=16$$

Commented by pieroo last updated on 18/Feb/19

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

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