Number of words that can be made using all the letters of the word W, if Os as well as Is are separated is?












8












$begingroup$


enter image description here



I am facing difficulty in solving the 18th Question from the above passage.



Attempt:



I have attempted it using principal of inclusion and exclusion.



Let n be the required number of ways.



$n = text{Total number of ways - Number of ways in which Os and Is are together}$



$implies n = dfrac{12!}{3!2!2!} - dfrac{11!}{2!3!} - dfrac{10!}{2!2!}+ dfrac{9!2!}{2!} = 399 times 8!$



This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?






combinatorics permutations combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    Mar 28 at 8:29










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    Mar 28 at 8:30
















8












$begingroup$


enter image description here



I am facing difficulty in solving the 18th Question from the above passage.



Attempt:



I have attempted it using principal of inclusion and exclusion.



Let n be the required number of ways.



$n = text{Total number of ways - Number of ways in which Os and Is are together}$



$implies n = dfrac{12!}{3!2!2!} - dfrac{11!}{2!3!} - dfrac{10!}{2!2!}+ dfrac{9!2!}{2!} = 399 times 8!$



This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?






combinatorics permutations combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    Mar 28 at 8:29










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    Mar 28 at 8:30














8












8








8


1



$begingroup$


enter image description here



I am facing difficulty in solving the 18th Question from the above passage.



Attempt:



I have attempted it using principal of inclusion and exclusion.



Let n be the required number of ways.



$n = text{Total number of ways - Number of ways in which Os and Is are together}$



$implies n = dfrac{12!}{3!2!2!} - dfrac{11!}{2!3!} - dfrac{10!}{2!2!}+ dfrac{9!2!}{2!} = 399 times 8!$



This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?






combinatorics permutations combinations







share|cite|improve this question









$endgroup$




enter image description here



I am facing difficulty in solving the 18th Question from the above passage.



Attempt:



I have attempted it using principal of inclusion and exclusion.



Let n be the required number of ways.



$n = text{Total number of ways - Number of ways in which Os and Is are together}$



$implies n = dfrac{12!}{3!2!2!} - dfrac{11!}{2!3!} - dfrac{10!}{2!2!}+ dfrac{9!2!}{2!} = 399 times 8!$



This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?






combinatorics permutations combinations




combinatorics permutations combinations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 28 at 7:09









AbcdAbcd

3,16931339




3,16931339








  • 1




    $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    Mar 28 at 8:29










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    Mar 28 at 8:30














  • 1




    $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    Mar 28 at 8:29










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    Mar 28 at 8:30








1




1




$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
Mar 28 at 8:29




$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
Mar 28 at 8:29












$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
Mar 28 at 8:30




$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
Mar 28 at 8:30










3 Answers
3






active

oldest

votes


















13












$begingroup$

Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.



One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.



With this correction, the final calculation becomes
$$
overbrace{frac{12!}{3!2!2!}}^{text{All words}} - overbrace{frac{11!}{3!2!}}^{text{O's not separate}} - overbrace{left(underbrace{frac{11!}{2!2!}}_{text{II and I}} - underbrace{frac{10!}{2!2!}}_{text{III}}right)}^{text{I's not separate}} + overbrace{left(underbrace{frac{10!}{2!}}_{text{OO, II and I}} - underbrace{frac{9!}{2!}}_{text{OO and III}}right)}^{text{Neither O's nor I's separate}}
$$
which turns out to be $8!cdot 228$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
    $endgroup$
    – Abcd
    Mar 28 at 9:21












  • $begingroup$
    @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
    $endgroup$
    – Arthur
    Mar 28 at 9:32








  • 1




    $begingroup$
    +1 for the MathJax magic!
    $endgroup$
    – TonyK
    Mar 28 at 12:09



















3












$begingroup$

The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    How did you get each of the terms of $n$?




    • $frac{12!}{3!2!2!}$ : The number of ways of arranging the letters of SOLICITATION.


    • $frac{11!}{2!3!}$ : The number of ways in which the two $O$s can be together in SOLICITATION.


    • $frac{10!}{2!2!}$ : The number of ways in which the three $I$s can be together in SOLICITATION.


    • $frac{9!2!}{2!}$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.



    Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      13












      $begingroup$

      Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.



      One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.



      With this correction, the final calculation becomes
      $$
      overbrace{frac{12!}{3!2!2!}}^{text{All words}} - overbrace{frac{11!}{3!2!}}^{text{O's not separate}} - overbrace{left(underbrace{frac{11!}{2!2!}}_{text{II and I}} - underbrace{frac{10!}{2!2!}}_{text{III}}right)}^{text{I's not separate}} + overbrace{left(underbrace{frac{10!}{2!}}_{text{OO, II and I}} - underbrace{frac{9!}{2!}}_{text{OO and III}}right)}^{text{Neither O's nor I's separate}}
      $$
      which turns out to be $8!cdot 228$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
        $endgroup$
        – Abcd
        Mar 28 at 9:21












      • $begingroup$
        @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
        $endgroup$
        – Arthur
        Mar 28 at 9:32








      • 1




        $begingroup$
        +1 for the MathJax magic!
        $endgroup$
        – TonyK
        Mar 28 at 12:09
















      13












      $begingroup$

      Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.



      One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.



      With this correction, the final calculation becomes
      $$
      overbrace{frac{12!}{3!2!2!}}^{text{All words}} - overbrace{frac{11!}{3!2!}}^{text{O's not separate}} - overbrace{left(underbrace{frac{11!}{2!2!}}_{text{II and I}} - underbrace{frac{10!}{2!2!}}_{text{III}}right)}^{text{I's not separate}} + overbrace{left(underbrace{frac{10!}{2!}}_{text{OO, II and I}} - underbrace{frac{9!}{2!}}_{text{OO and III}}right)}^{text{Neither O's nor I's separate}}
      $$
      which turns out to be $8!cdot 228$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
        $endgroup$
        – Abcd
        Mar 28 at 9:21












      • $begingroup$
        @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
        $endgroup$
        – Arthur
        Mar 28 at 9:32








      • 1




        $begingroup$
        +1 for the MathJax magic!
        $endgroup$
        – TonyK
        Mar 28 at 12:09














      13












      13








      13





      $begingroup$

      Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.



      One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.



      With this correction, the final calculation becomes
      $$
      overbrace{frac{12!}{3!2!2!}}^{text{All words}} - overbrace{frac{11!}{3!2!}}^{text{O's not separate}} - overbrace{left(underbrace{frac{11!}{2!2!}}_{text{II and I}} - underbrace{frac{10!}{2!2!}}_{text{III}}right)}^{text{I's not separate}} + overbrace{left(underbrace{frac{10!}{2!}}_{text{OO, II and I}} - underbrace{frac{9!}{2!}}_{text{OO and III}}right)}^{text{Neither O's nor I's separate}}
      $$
      which turns out to be $8!cdot 228$.






      share|cite|improve this answer











      $endgroup$



      Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.



      One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.



      With this correction, the final calculation becomes
      $$
      overbrace{frac{12!}{3!2!2!}}^{text{All words}} - overbrace{frac{11!}{3!2!}}^{text{O's not separate}} - overbrace{left(underbrace{frac{11!}{2!2!}}_{text{II and I}} - underbrace{frac{10!}{2!2!}}_{text{III}}right)}^{text{I's not separate}} + overbrace{left(underbrace{frac{10!}{2!}}_{text{OO, II and I}} - underbrace{frac{9!}{2!}}_{text{OO and III}}right)}^{text{Neither O's nor I's separate}}
      $$
      which turns out to be $8!cdot 228$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 28 at 9:32

























      answered Mar 28 at 8:37









      ArthurArthur

      121k7122208




      121k7122208












      • $begingroup$
        Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
        $endgroup$
        – Abcd
        Mar 28 at 9:21












      • $begingroup$
        @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
        $endgroup$
        – Arthur
        Mar 28 at 9:32








      • 1




        $begingroup$
        +1 for the MathJax magic!
        $endgroup$
        – TonyK
        Mar 28 at 12:09


















      • $begingroup$
        Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
        $endgroup$
        – Abcd
        Mar 28 at 9:21












      • $begingroup$
        @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
        $endgroup$
        – Arthur
        Mar 28 at 9:32








      • 1




        $begingroup$
        +1 for the MathJax magic!
        $endgroup$
        – TonyK
        Mar 28 at 12:09
















      $begingroup$
      Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
      $endgroup$
      – Abcd
      Mar 28 at 9:21






      $begingroup$
      Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
      $endgroup$
      – Abcd
      Mar 28 at 9:21














      $begingroup$
      @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
      $endgroup$
      – Arthur
      Mar 28 at 9:32






      $begingroup$
      @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
      $endgroup$
      – Arthur
      Mar 28 at 9:32






      1




      1




      $begingroup$
      +1 for the MathJax magic!
      $endgroup$
      – TonyK
      Mar 28 at 12:09




      $begingroup$
      +1 for the MathJax magic!
      $endgroup$
      – TonyK
      Mar 28 at 12:09











      3












      $begingroup$

      The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.






          share|cite|improve this answer









          $endgroup$



          The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 28 at 8:38









          jmerryjmerry

          16.9k11633




          16.9k11633























              3












              $begingroup$

              How did you get each of the terms of $n$?




              • $frac{12!}{3!2!2!}$ : The number of ways of arranging the letters of SOLICITATION.


              • $frac{11!}{2!3!}$ : The number of ways in which the two $O$s can be together in SOLICITATION.


              • $frac{10!}{2!2!}$ : The number of ways in which the three $I$s can be together in SOLICITATION.


              • $frac{9!2!}{2!}$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.



              Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                How did you get each of the terms of $n$?




                • $frac{12!}{3!2!2!}$ : The number of ways of arranging the letters of SOLICITATION.


                • $frac{11!}{2!3!}$ : The number of ways in which the two $O$s can be together in SOLICITATION.


                • $frac{10!}{2!2!}$ : The number of ways in which the three $I$s can be together in SOLICITATION.


                • $frac{9!2!}{2!}$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.



                Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  How did you get each of the terms of $n$?




                  • $frac{12!}{3!2!2!}$ : The number of ways of arranging the letters of SOLICITATION.


                  • $frac{11!}{2!3!}$ : The number of ways in which the two $O$s can be together in SOLICITATION.


                  • $frac{10!}{2!2!}$ : The number of ways in which the three $I$s can be together in SOLICITATION.


                  • $frac{9!2!}{2!}$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.



                  Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.






                  share|cite|improve this answer









                  $endgroup$



                  How did you get each of the terms of $n$?




                  • $frac{12!}{3!2!2!}$ : The number of ways of arranging the letters of SOLICITATION.


                  • $frac{11!}{2!3!}$ : The number of ways in which the two $O$s can be together in SOLICITATION.


                  • $frac{10!}{2!2!}$ : The number of ways in which the three $I$s can be together in SOLICITATION.


                  • $frac{9!2!}{2!}$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.



                  Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 28 at 8:51









                  астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                  40.2k33577




                  40.2k33577






























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                      Puebla de Zaragoza

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                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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