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












4












$begingroup$


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







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$endgroup$












  • $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    1 hour ago










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    1 hour ago
















4












$begingroup$


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









$endgroup$












  • $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    1 hour ago










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    1 hour ago














4












4








4





$begingroup$


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 2 hours ago









AbcdAbcd

3,08931337




3,08931337












  • $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    1 hour ago










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    1 hour ago


















  • $begingroup$
    How do you know that your solution is wrong? There is also option (D)
    $endgroup$
    – Robert Z
    1 hour ago










  • $begingroup$
    @RobertZ answer given is option B
    $endgroup$
    – Abcd
    1 hour ago
















$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
1 hour ago




$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
1 hour ago












$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
1 hour ago




$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
1 hour ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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
    32 mins ago












  • $begingroup$
    @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
    21 mins ago





















2












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      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
        32 mins ago












      • $begingroup$
        @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
        21 mins ago


















      4












      $begingroup$

      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
        32 mins ago












      • $begingroup$
        @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
        21 mins ago
















      4












      4








      4





      $begingroup$

      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 21 mins ago

























      answered 1 hour ago









      ArthurArthur

      120k7120204




      120k7120204












      • $begingroup$
        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
        32 mins ago












      • $begingroup$
        @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
        21 mins ago




















      • $begingroup$
        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
        32 mins ago












      • $begingroup$
        @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
        21 mins ago


















      $begingroup$
      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
      32 mins ago






      $begingroup$
      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
      32 mins ago














      $begingroup$
      @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
      21 mins ago






      $begingroup$
      @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
      21 mins ago













      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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 1 hour ago









          jmerryjmerry

          16.6k11633




          16.6k11633























              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 1 hour ago









                  астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                  40k33477




                  40k33477






























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                      Bunad

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                      Tradisjonelle folkedrakter frå Aust-Telemark, truleg frå 1880-åra. Drakter som dette er grunnlaget for moderne bunadar. Foto: Axel Lindahl Bunad frå Eggedal i Buskerud fotografert i 1906. Bunad (frå kledebunad , der norrønt «búnaðr» har same rot som (føre)bu) er ei samnemning brukt om klesdrakter i tradisjonell nordisk stil — ofte utvikla frå folkedrakter, men likevel i ein eigen tradisjon. Ein bunad er typisk ei forseggjort og kostbar drakt, med mykje broderi, og med handsmidde smykke i gull eller sølv, særleg dei særeigne søljene, som obligatorisk prynad i tillegg. Bunad og bunadbruk er eit debattemne med mange ulike oppfatningar, til dømes om kva som er ein bunad, og om ein kan endra på utforminga av draktene eller ikkje. Den norske bunadrørsla går tilbake til folkedansrørsla og føregangsfolk som Hulda Garborg og Klara Semb. I dag blir bunad brukt som nasjonaldrakt. Festbunad er likestilt med galla-antrekk, og ofte nytta ved særskilde høve, som konfirmasjonar eller d
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