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How-many-different-words-can-be-formed-with-the-same-letters-as-in-TINKUTARA-if-no-two-same-letters-are-next-to-each-other-

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Question Number 107196 by mr W last updated on 09/Aug/20
How many different words can be formed with the same letters as in TINKUTARA if no two same letters are next to each other.
HowmanydifferentwordscanbeformedwiththesamelettersasinTINKUTARAifnotwosamelettersarenexttoeachother.
Commented by john santu last updated on 09/Aug/20
totally word = ((9!)/(2!.2!)) case(1) □□□□− − − − − = 5! −□□□□− − − −= 5! − −□□□□− − − = 5! − − −□□□□− − = 5! − − − −□□□□−= 5! − − − − −□□□□ = 5! ___________________ + = 2×6×5 ! note □□=AA & □□=TT case(2) □□−□□− − − −= 5! □□− −□□− − − = 5! □□− − −□□− −=5! □□− − − −□□−=5! □□− − − − −□□=5! ________________ totall = 2×5×5! case(3) −□□−□□− − −=5! ⇒total=2×4×5! case(4) − −□□−□□− −=5!⇒total=2×3×5! case(5) − − −□□−□□−=5!⇒total=2×2×5! therefore many different word can be formed with the same letters as in Tinkutara if no two same letters are next to each other = ((9!)/(2!.2!)) − 2×5!(6+5+4+3+2) = ((9!)/(2!.2!))−40.5!
totallyword=9!2!.2!case(1)◻◻◻◻−−−−−=5!−◻◻◻◻−−−−=5!−−◻◻◻◻−−−=5!−−−◻◻◻◻−−=5!−−−−◻◻◻◻−=5!−−−−−◻◻◻◻=5!___________________+=2×6×5!note◻◻=AA&◻◻=TTcase(2)◻◻−◻◻−−−−=5!◻◻−−◻◻−−−=5!◻◻−−−◻◻−−=5!◻◻−−−−◻◻−=5!◻◻−−−−−◻◻=5!________________totall=2×5×5!case(3)−◻◻−◻◻−−−=5!⇒total=2×4×5!case(4)−−◻◻−◻◻−−=5!⇒total=2×3×5!case(5)−−−◻◻−◻◻−=5!⇒total=2×2×5!thereforemanydifferentwordcanbeformedwiththesamelettersasinTinkutaraifnotwosamelettersarenexttoeachother=9!2!.2!−2×5!(6+5+4+3+2)=9!2!.2!−40.5!
Commented by mr W last updated on 09/Aug/20
i got 55440.
igot55440.
Commented by mr W last updated on 09/Aug/20
but −□□−□−−□− is also not valid.
but−◻◻−◻−−◻−isalsonotvalid.
Commented by bobhans last updated on 09/Aug/20
case(6) □□□− □− − − −=5!⇒total=2×5×5! case(7) □□−−□−□−−=5!⇒total=2×3×5! next...
case(6)◻◻◻−◻−−−−=5!⇒total=2×5×5!case(7)◻◻−−◻−◻−−=5!⇒total=2×3×5!next…
Commented by bobhans last updated on 09/Aug/20
what the formula?
whattheformula?
Commented by mr W last updated on 09/Aug/20
C_2 ^8 ×((7!)/(2!))−C_2 ^7 ×6!=55440
C28×7!2!−C27×6!=55440
Commented by bobhans last updated on 09/Aug/20
C_2 ^8 ×((7!)/(2!)) = ((8.7)/(2.1))×((7!)/(2!)) = ((8!.7)/(2!.2!)) why not same to ((9!)/(2!.2!)) ? Tinkutara = 9 object where→ { ((A=2)),((T=2)) :} ?
C28×7!2!=8.72.1×7!2!=8!.72!.2!whynotsameto9!2!.2!?Tinkutara=9objectwhere→{A=2T=2?
Commented by mr W last updated on 09/Aug/20
i used a different method to solve.
iusedadifferentmethodtosolve.
Answered by mr W last updated on 09/Aug/20
METHOD I we have totally 9 letters, among them two letters T and two letters A. both Ts and As should not be next to each other. let′s consider at first that only the two Ts should not be next to each other. that means the two Ts must be separated by at least one of the other 7 letters: □■□■□■□■□■□■□■□ wherein ■ the 7 other letters (AINKURA) □ possible places for the two Ts to select two places for Ts there are C_2 ^8 ways and to arrange the 7 other letters there are ((7!)/(2!)) ways, with 2! because we have two same letters A. so we can form C_2 ^8 ×((7!)/(2!)) words without Ts next to each other. in next step let′s see how many words have both letters A next to each other. we consider the two letters A as a single letter, so we have only 6 other letters instead of 7: □■□■□■□■□■□■□ wherein ■ the 6 other letters (INKURA with A=AA as fixed unit) □ possible places for the two Ts to select two places for Ts there are C_2 ^7 ways and to arrange the 6 other letters there are 6! ways. so we can form C_2 ^7 ×6! words without Ts next to each other but with two As next to each other. number of words without Ts next to each other and without As next to each other is: C_2 ^8 ×((7!)/(2!))−C_2 ^7 ×6!=55440
METHODIwehavetotally9letters,amongthemtwolettersTandtwolettersA.bothTsandAsshouldnotbenexttoeachother.let′sconsideratfirstthatonlythetwoTsshouldnotbenexttoeachother.thatmeansthetwoTsmustbeseparatedbyatleastoneoftheother7letters:◻◼◻◼◻◼◻◼◻◼◻◼◻◼◻wherein◼the7otherletters(AINKURA)◻possibleplacesforthetwoTstoselecttwoplacesforTsthereareC28waysandtoarrangethe7otherlettersthereare7!2!ways,with2!becausewehavetwosamelettersA.sowecanformC28×7!2!wordswithoutTsnexttoeachother.innextsteplet′sseehowmanywordshavebothlettersAnexttoeachother.weconsiderthetwolettersAasasingleletter,sowehaveonly6otherlettersinsteadof7:◻◼◻◼◻◼◻◼◻◼◻◼◻wherein◼the6otherletters(INKURAwithA=AAasfixedunit)◻possibleplacesforthetwoTstoselecttwoplacesforTsthereareC27waysandtoarrangethe6otherlettersthereare6!ways.sowecanformC27×6!wordswithoutTsnexttoeachotherbutwithtwoAsnexttoeachother.numberofwordswithoutTsnexttoeachotherandwithoutAsnexttoeachotheris:C28×7!2!−C27×6!=55440
Answered by mr W last updated on 09/Aug/20
METHOD II total: ((9!)/(2!2!)) two A and two T together: XXTTXXAAX 7! two A or two T together: XXAAXXTXT XXTTXXAXA 2×(((8!)/(2!))−7!) result: ((9!)/(2!2!))−7!−2×(((8!)/(2!))−7!) =((9!)/(2!2!))−8!+7!=55440
METHODIItotal:9!2!2!twoAandtwoTtogether:XXTTXXAAX7!twoAortwoTtogether:XXAAXXTXTXXTTXXAXA2×(8!2!−7!)result:9!2!2!−7!−2×(8!2!−7!)=9!2!2!−8!+7!=55440

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