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Question Number 218076 by malwan last updated on 28/Mar/25

How many ways to arrnge  the letters ABCCCDEFG  (1) in general .  (2) all 3 Cs must be together  (3) only 2 Cs must be together  (4) no 2 or 3 Cs be together  (5) no letter still  in its  original place .

HowmanywaystoarrngethelettersABCCCDEFG(1)ingeneral.(2)all3Csmustbetogether(3)only2Csmustbetogether(4)no2or3Csbetogether(5)noletterstillinitsoriginalplace.

Answered by efronzo1 last updated on 29/Mar/25

(1) ((9!)/(3!))   (2) 7!

(1)9!3!(2)7!

Answered by mr W last updated on 29/Mar/25

(1)  ((9!)/(3!))=60480    (2)  7!=5040   or 6!×7=5040    (3)  □A□B□D□E□F□G□  6!×7×6=30240  the two Cs and a C must be separated   by other letters.  they can be placed   into the 7 red boxes.  to arrange the 6 other letters, there  are 6! ways, to place two Cs and a C,  there are 7×6 ways.  totally 6!×7×6    (4)  6!×7×6×5/3!=25200  see also above. the three Cs must  be separated by the other letters.  to place the three Cs into the 7  boxes, there are C_7 ^3  =7×6×5/3!  ways.  totally 6!×7×6×5/3!    (5)  ABCCCDEFG  EFABDGCCC  C_6 ^3 ×3!×(((6!)/(3!))−1−3×3−3×13)=8520  under the three Cs other letters  must be placed. to select 3 letters  and place them under the Cs there  are C_6 ^3 ×3!=120 ways. say  A,B,D are selected. to arrange the  remaining 6 letters (eg. EFGCCC)  there are totally ((6!)/(3!)) ways. but there  are arrangements in which EFG  are placed under themselves.  all three letters (EFG) are in false  positions: there is one possibility.  two letters (eg. EF) are in false  positions: there are 3×4 possibilities.  but this also includes 3 times the  case from above. that means there  are 3×3 possibilities for only two  letters in false positions.  one letter (eg. E) in false position:  there are 3×5×4 possibilities. but  this also includes 7 times the   cases above. that means there are  3×13 possibilities for only one  letter in false position.  totally C_6 ^3 ×3!×(((6!)/(3!))−1−3×3−3×13)

(1)9!3!=60480(2)7!=5040or6!×7=5040(3)ABDEFG6!×7×6=30240thetwoCsandaCmustbeseparatedbyotherletters.theycanbeplacedintothe7redboxes.toarrangethe6otherletters,thereare6!ways,toplacetwoCsandaC,thereare7×6ways.totally6!×7×6(4)6!×7×6×5/3!=25200seealsoabove.thethreeCsmustbeseparatedbytheotherletters.toplacethethreeCsintothe7boxes,thereareC73=7×6×5/3!ways.totally6!×7×6×5/3!(5)ABCCCDEFGEFABDGCCCC63×3!×(6!3!13×33×13)=8520underthethreeCsotherlettersmustbeplaced.toselect3lettersandplacethemundertheCsthereareC63×3!=120ways.sayA,B,Dareselected.toarrangetheremaining6letters(eg.EFGCCC)therearetotally6!3!ways.buttherearearrangementsinwhichEFGareplacedunderthemselves.allthreeletters(EFG)areinfalsepositions:thereisonepossibility.twoletters(eg.EF)areinfalsepositions:thereare3×4possibilities.butthisalsoincludes3timesthecasefromabove.thatmeansthereare3×3possibilitiesforonlytwolettersinfalsepositions.oneletter(eg.E)infalseposition:thereare3×5×4possibilities.butthisalsoincludes7timesthecasesabove.thatmeansthereare3×13possibilitiesforonlyoneletterinfalseposition.totallyC63×3!×(6!3!13×33×13)

Commented by malwan last updated on 29/Mar/25

thanks alot , mr W

thanksalot,mrW

Commented by mr W last updated on 29/Mar/25

i′ve added some explanation.

iveaddedsomeexplanation.

Commented by malwan last updated on 29/Mar/25

your answer maybe like this  ((6!)/(3!))×(((6!)/(3!))−1−3×3−3×17+12)  = 8520  but , why +12 ?  I hope that you will find it  thanks again  ⋛

youranswermaybelikethis6!3!×(6!3!13×33×17+12)=8520but,why+12?Ihopethatyouwillfinditthanksagain

Commented by mr W last updated on 29/Mar/25

answer for (5) can also be determined  applying “rook polynomial”  ((51120)/(3!))=8520

answerfor(5)canalsobedeterminedapplyingrookpolynomial511203!=8520

Commented by mr W last updated on 29/Mar/25

Commented by mr W last updated on 29/Mar/25

i′ve got the right answer.  it is −3×13, not −3×17.

ivegottherightanswer.itis3×13,not3×17.

Commented by malwan last updated on 29/Mar/25

thank you so much sir

thankyousomuchsir

Commented by mr W last updated on 30/Mar/25

this is the explanation for why   −3×13:  say we have placed ABD under CCC  ABCCCDEFG  □□ABD□□□□  now we try to place EFGCCC into  the 6 boxes. totally there are ((6!)/(3!))   ways. but they include also following  cases:  1) all EFG in false positions  CCABDCEFG  ⇒just one time  2) only two of EFG in false positions  i.e. EG or EG or FG in false positions  say EG in false positions:  □□ABD□EF■  G can be in white boxes, 3 possibilities.  totally 3×3 possibilities.  3) only one of EFG in false position  i.e. E or F or G in false position  say E in false position:  □□ABD□E□□  there are 5×4=20 possibilities  to place FG and Cs. but this includes   □□ABD□EFG ⇒1 time  □□ABD□EF■ ⇒3 times (G in □)  □□ABD□E■G ⇒3 times (F in □)  therefore we have 20−7=13 cases  with only E in false position.

thisistheexplanationforwhy3×13:saywehaveplacedABDunderCCCABCCCDEFGABDnowwetrytoplaceEFGCCCintothe6boxes.totallythereare6!3!ways.buttheyincludealsofollowingcases:1)allEFGinfalsepositionsCCABDCEFGjustonetime2)onlytwoofEFGinfalsepositionsi.e.EGorEGorFGinfalsepositionssayEGinfalsepositions:ABDEFGcanbeinwhiteboxes,3possibilities.totally3×3possibilities.3)onlyoneofEFGinfalsepositioni.e.EorForGinfalsepositionsayEinfalseposition:ABDEthereare5×4=20possibilitiestoplaceFGandCs.butthisincludesABDEFG1timeABDEF3times(Gin)ABDEG3times(Fin)thereforewehave207=13caseswithonlyEinfalseposition.

Commented by malwan last updated on 31/Mar/25

Thank you trillion  ⋛

Thankyoutrillion

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