No boy stands next to each others permutation problem












6












$begingroup$


The question is:




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




In which the correct model would be:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
$$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update

Also, if we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
What is wrong with the last two logic?

Thank you.






combinatorics permutations







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    The question is:




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




    In which the correct model would be:
    $$square g square g square g square g square g square gsquare$$
    Therefore:
    $$6!cdot 7P5=1space 814space 400$$
    However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
    $$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
    We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
    $$5!cdot6P5cdot12=1space036space800$$
    Update

    Also, if we are to swap it around:
    $$square b square b square b square b square b square$$
    We now have:
    $$5!cdot6!=86space 400 $$
    What is wrong with the last two logic?

    Thank you.






    combinatorics permutations







    share|cite|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$


      The question is:




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




      In which the correct model would be:
      $$square g square g square g square g square g square gsquare$$
      Therefore:
      $$6!cdot 7P5=1space 814space 400$$
      However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
      $$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
      We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
      $$5!cdot6P5cdot12=1space036space800$$
      Update

      Also, if we are to swap it around:
      $$square b square b square b square b square b square$$
      We now have:
      $$5!cdot6!=86space 400 $$
      What is wrong with the last two logic?

      Thank you.






      combinatorics permutations







      share|cite|improve this question











      $endgroup$




      The question is:




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




      In which the correct model would be:
      $$square g square g square g square g square g square gsquare$$
      Therefore:
      $$6!cdot 7P5=1space 814space 400$$
      However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
      $$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
      We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
      $$5!cdot6P5cdot12=1space036space800$$
      Update

      Also, if we are to swap it around:
      $$square b square b square b square b square b square$$
      We now have:
      $$5!cdot6!=86space 400 $$
      What is wrong with the last two 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 12 mins ago







      Huy Tran Van

















      asked 2 hours ago









      Huy Tran VanHuy Tran Van

      314




      314






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          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
            1 hour ago












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












          • $begingroup$
            I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
            $endgroup$
            – Huy Tran Van
            46 mins ago





















          2












          $begingroup$

          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












            $begingroup$

            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
              1 hour ago











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084008%2fno-boy-stands-next-to-each-others-permutation-problem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            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
              1 hour ago












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












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              46 mins ago


















            3












            $begingroup$

            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
              1 hour ago












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












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              46 mins ago
















            3












            3








            3





            $begingroup$

            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 1 hour ago









            orlporlp

            7,4211330




            7,4211330












            • $begingroup$
              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
              1 hour ago












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












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              46 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
              1 hour ago












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












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              46 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
            1 hour 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
            1 hour ago














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






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














            $begingroup$
            I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
            $endgroup$
            – Huy Tran Van
            46 mins ago






            $begingroup$
            I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
            $endgroup$
            – Huy Tran Van
            46 mins ago













            2












            $begingroup$

            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











            $endgroup$


















              2












              $begingroup$

              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











              $endgroup$
















                2












                2








                2





                $begingroup$

                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 1 hour ago

























                answered 1 hour ago









                Chris CusterChris Custer

                11.5k3824




                11.5k3824























                    -1












                    $begingroup$

                    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
                      1 hour ago
















                    -1












                    $begingroup$

                    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
                      1 hour ago














                    -1












                    -1








                    -1





                    $begingroup$

                    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 1 hour 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
                      1 hour 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
                      1 hour 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
                    1 hour 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
                    1 hour ago


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084008%2fno-boy-stands-next-to-each-others-permutation-problem%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    What other Star Trek series did the main TNG cast show up in?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75 ...
                    Read more

                    Berlina muro

                    Image
                    La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l...
                    Read more

                    Berlina aerponto

                    Image
                    Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa...
                    Read more