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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                      In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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                      How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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                      1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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