Todeterminehowmanydistinct4digitnumberscanbeformedusingthedigits1,2,3,and4withoutrepetition,weanalyzetheproblemasfollows:

StepbyStepReasoning

˙▂˙ Wearegivenfourdistinctdigits:1,2,3,and4.Wewanttoform4digitnumbers,meaningallfourdigitsmustbeusedexactlyonceineachnumber.

Thisisaclassicpermutationproblemwhere:

≥﹏≤ Alldigitsareused.

Nodigitisrepeated.

Theorderofdigitsmatters(since1234≠1243).

?０? Thetotalnumberofsuchpermutationsiscalculatedusingtheformulaforpermutationsof$n$itemstaken$r$atatime:

$$

P(n,r)=\frac{n!}{(nr)!}

$$

Here,$n=4$and$r=4$,so:

$$

P(4,4)=\frac{4!}{(44)!}=\frac{4!}{0!}=\frac{24}{1}=24

$$

Alternatively,wecancomputeitdirectlybyconsideringthechoicesavailableforeachposition:

ˋωˊ Firstdigit:4choices

Seconddigit:3remainingchoices

Thirddigit:2remainingchoices

Fourthdigit:1remainingchoice

$$

\text{Totalcombinations}=4\times3\times2\times1=24

$$

FinalAnswer

Allresultingnumbersarevalid4digitnumberssincenoneofthedigitsiszero,andeverypermutationusesallfourdigitsexactlyonce.Therefore,therearenoinvalidcasesduetoleadingzerosorotherconstraints.

$$

ˋ▽ˊ \boxed{24}

$$