Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy












5












$begingroup$


The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



$$square g square g square g square g square g square gsquare$$
Therefore, we come up with:



$$6!cdot 7P5=1space 814space 400$$



However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



$$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
$$5!cdot6P5cdot12=1space036space800$$



Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
$$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



$$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
In which we have:
$$2(5!)^2cdot6=172space800$$



So which is right? Is there anything wrong with each logic?



Thank you.






combinatorics permutations







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



    $$square g square g square g square g square g square gsquare$$
    Therefore, we come up with:



    $$6!cdot 7P5=1space 814space 400$$



    However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



    $$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



    We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
    $$5!cdot6P5cdot12=1space036space800$$



    Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
    $$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



    Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



    $$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
    In which we have:
    $$2(5!)^2cdot6=172space800$$



    So which is right? Is there anything wrong with each logic?



    Thank you.






    combinatorics permutations







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



      $$square g square g square g square g square g square gsquare$$
      Therefore, we come up with:



      $$6!cdot 7P5=1space 814space 400$$



      However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



      $$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



      We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
      $$5!cdot6P5cdot12=1space036space800$$



      Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
      $$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



      Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



      $$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
      In which we have:
      $$2(5!)^2cdot6=172space800$$



      So which is right? Is there anything wrong with each logic?



      Thank you.






      combinatorics permutations







      share|cite|improve this question











      $endgroup$




      The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



      $$square g square g square g square g square g square gsquare$$
      Therefore, we come up with:



      $$6!cdot 7P5=1space 814space 400$$



      However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



      $$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



      We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
      $$5!cdot6P5cdot12=1space036space800$$



      Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
      $$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



      Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



      $$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
      In which we have:
      $$2(5!)^2cdot6=172space800$$



      So which is right? Is there anything wrong with each logic?



      Thank you.






      combinatorics permutations




      combinatorics permutations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 42 mins ago







      Huy Tran Van

















      asked 1 hour ago









      Huy Tran VanHuy Tran Van

      264




      264






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Your first computation is correct.



          The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



          Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
            $endgroup$
            – Huy Tran Van
            16 mins ago












          • $begingroup$
            @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
            $endgroup$
            – orlp
            5 mins ago





















          1












          $begingroup$

          In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



          In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



          The right one is the first.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

            A $7 choose 5$ is calculated as
            $$
            frac{7!}{5!*(7-5)!} = 21
            $$
            Therefore there are 21 combinations in which the boys could be arranged.



            I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



            Picture of how the boys and girls could be arranged in 21 combinations






            share|cite|improve this answer








            New contributor




            Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
              $endgroup$
              – Tucktuckgoose
              29 mins ago











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              16 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              5 mins ago


















            2












            $begingroup$

            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              16 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              5 mins ago
















            2












            2








            2





            $begingroup$

            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






            share|cite|improve this answer









            $endgroup$



            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 41 mins ago









            orlporlp

            7,4111330




            7,4111330












            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              16 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              5 mins ago




















            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              16 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              5 mins ago


















            $begingroup$
            Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
            $endgroup$
            – Huy Tran Van
            16 mins ago






            $begingroup$
            Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
            $endgroup$
            – Huy Tran Van
            16 mins ago














            $begingroup$
            @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
            $endgroup$
            – orlp
            5 mins ago






            $begingroup$
            @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
            $endgroup$
            – orlp
            5 mins ago













            1












            $begingroup$

            In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



            In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



            The right one is the first.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



              In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



              The right one is the first.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



                In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



                The right one is the first.






                share|cite|improve this answer











                $endgroup$



                In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



                In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



                The right one is the first.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 34 mins ago

























                answered 41 mins ago









                Chris CusterChris Custer

                11.5k3824




                11.5k3824























                    0












                    $begingroup$

                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations






                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      29 mins ago
















                    0












                    $begingroup$

                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations






                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      29 mins ago














                    0












                    0








                    0





                    $begingroup$

                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations






                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations







                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 32 mins ago









                    TucktuckgooseTucktuckgoose

                    11




                    11




                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.












                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      29 mins ago


















                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      29 mins ago
















                    $begingroup$
                    My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                    $endgroup$
                    – Tucktuckgoose
                    29 mins ago




                    $begingroup$
                    My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                    $endgroup$
                    – Tucktuckgoose
                    29 mins ago


















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                    Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове
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                    Josef Freinademetz

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                    San Ujöp (Giuseppe/Josef) Freinademetz Josef Freinademetz in abiti cinesi   Religioso e missionario   Nascita 15 aprile 1852 Morte 28 gennaio 1908 Venerato da Chiesa cattolica Beatificazione 1975 Canonizzazione 5 ottobre 2003 Ricorrenza 28 gennaio Manuale Ujöp Freinademetz noto anche come Giuseppe o Josef Freinademetz (Oies, 15 aprile 1852 – Taikia, 28 gennaio 1908) è stato un presbitero austriaco, membro della Società del Verbo Divino, fu missionario in Cina; è venerato come santo dalla Chiesa cattolica. .mw-parser-output .citazione-table{margin-bottom:.5em;font-size:95%}.mw-parser-output .citazione-table td{padding:0 1.2em 0 2.4em}.mw-parser-output .citazione-lang{vertical-align:top}.mw-parser-output .citazione-lang td{width:50%}.mw-parser-output .citazione-lang td:first-child{padding:0 0 0 2.4em}.mw-parser-output .citazione-lang td:nth-child(2){padding:0 1.2em} «Io amo la Cina e i cinesi; voglio morire in mezzo a loro, e tra loro
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