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
              30 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
                      30 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
                      30 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
                      30 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
                      30 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
                    30 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
                    30 mins ago


















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                    Tanganjiko

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                    Lago Tanganjiko Situo centra Afriko Ĉefaj fontoj Kalambo, Lufuko, Malagarasi kaj Ruzizi Ĉefaj elfluoj Lukugo Setlejoj Bujumbura, Kalemie Datenoj Koordinatoj 6°  S , 30°  O -6.3827777777778 29.615 782 Koordinatoj: 6°  S , 30°  O DEC Supermara alteco 782 m Surfaca areo 32 893 km² f5 Maksimuma longo 673 km f6 Maksimuma larĝo 72 km f7 Akva volumeno 18.900 km³ dep1 f8 Maksimuma profundo 1470 m f10 Averaĝa profundo 570 m f11 v   •   d   •   r Mapo de la lago Tanganjiko estas lago en orienta Afriko, inter Burundo, Kongo Kinŝasa, Tanzanio kaj Zambio. Ĝi apartenas al la Granda Rifto kaj estas unu el la Afrikaj Grandaj Lagoj. Tanzanio (46%) kaj DRK (40%) havas la majoritaton de la lago. Ĝi havas longon de 673 km, areon de ĉ. 32.900 km² kaj maksimuman profundon taksatan je 1470 metroj; pro tiu ĉi granda profundo ankaŭ ĝia volumeno estas sufiĉe granda, 18.900 km³. Oni konsi
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                    Liste der Baudenkmäler in Enneberg

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                    Die Liste der Baudenkmäler in Enneberg (ladinisch Mareo , italienisch Marebbe ) enthält die 36 als Baudenkmäler ausgewiesenen Objekte auf dem Gebiet der Gemeinde Enneberg in Südtirol. Basis ist das im Internet einsehbare offizielle Verzeichnis der Baudenkmäler in Südtirol. Dabei kann es sich beispielsweise um Sakralbauten, Wohnhäuser, Bauernhöfe und Adelsansitze handeln. Die Reihenfolge in dieser Liste orientiert sich an der Bezeichnung, alternativ ist sie auch nach der Adresse oder dem Datum der Unterschutzstellung sortierbar. Liste | Foto   Bezeichnung Standort Eintragung Beschreibung Metadaten ja Altes Gasthaus Janesch ID: 14567 46° 42′ 54″ N, 11° 52′ 42″ O 76882557800 0 8. Sep. 1986 (BLR-LAB 4974) Ländliches Wohnhaus/Bauernhof KG:  Welschellen Bauparzelle: 391 ja Asch (Prack) ID: 14540 46° 43′ 12″ N, 11° 54′ 35″ O 75496485300 20. Nov. 1950 (MD) Ansitz KG:  Enneberg Bauparzelle: 38 ja Biccia mit Stade
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