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Permutation Combination And Probability Problems Pdf Free

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21400E….None….of….theseB…..60View..Answer….Discussanswer..with..explanationAnswer:..Option..CExplanation:..The..word..'BIHAR'..has..5..letters..and..all..these..5..letters..are..different.Total..number..of..words..that..can..be..formed..by..using..all..these..5..letters=..5P5..=5!=54321=12017…All..students,..freshers..can..download..Aptitude..Permutation..and..Combination..quiz..questions..with..answers..as..PDF..files..and..eBooks…6020C….

924000B…If….not….the….factorial….case….is….wrong…..6…There..is..only..1..way..of..doing..this.1Since..the..number..5..is..placed..at..unit..place,..we..have..now..five..digits(2,3,6,7,9)..remaining…825D…..9450View…Answer……Discussanswer…with…explanationAnswer:…Option…DExplanation:…Number…of…ways…to…choose…8…questions…from…part…P…=…10C8Number…of…ways…to…choose…4…questions…from…part…Q…=…10C4Total…number…of…ways=…10C8……10C4=…10C2……10C4[…nCr…=…nC(n-r)]=left(dfrac{109}{21}right)left(dfrac{10987}{4321}right)=45210=945028….277200C…that..is,..CRPRTN(OOAIO).Hence..we..can..assume..total..letters..as..7…5…..Each..subject..must..be..allowed..in..at..least..one..period…

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