How many different arrangements are there if Bob and Sally must always be seated next to each other?












0












$begingroup$


How many different arrangements are there in which Bob, Sally and $n$ other people sit down in a row of $n+3$ chairs if Bob and Sally must always be seated next to each other?



I tried putting Bob and Sally next to each other in the first two chairs so then there are $2 times n!$ arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this.






combinatorics permutations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:43










  • $begingroup$
    I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this
    $endgroup$
    – user607735
    Dec 7 '18 at 9:45






  • 1




    $begingroup$
    Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:47










  • $begingroup$
    Is the problem stated correctly, so one seat remains empty?
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:53










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 7 '18 at 10:07
















0












$begingroup$


How many different arrangements are there in which Bob, Sally and $n$ other people sit down in a row of $n+3$ chairs if Bob and Sally must always be seated next to each other?



I tried putting Bob and Sally next to each other in the first two chairs so then there are $2 times n!$ arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this.






combinatorics permutations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:43










  • $begingroup$
    I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this
    $endgroup$
    – user607735
    Dec 7 '18 at 9:45






  • 1




    $begingroup$
    Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:47










  • $begingroup$
    Is the problem stated correctly, so one seat remains empty?
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:53










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 7 '18 at 10:07














0












0








0





$begingroup$


How many different arrangements are there in which Bob, Sally and $n$ other people sit down in a row of $n+3$ chairs if Bob and Sally must always be seated next to each other?



I tried putting Bob and Sally next to each other in the first two chairs so then there are $2 times n!$ arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this.






combinatorics permutations







share|cite|improve this question











$endgroup$




How many different arrangements are there in which Bob, Sally and $n$ other people sit down in a row of $n+3$ chairs if Bob and Sally must always be seated next to each other?



I tried putting Bob and Sally next to each other in the first two chairs so then there are $2 times n!$ arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this.






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 Dec 7 '18 at 10:07









N. F. Taussig

44.1k93356




44.1k93356










asked Dec 7 '18 at 9:39









user607735user607735

103




103








  • 3




    $begingroup$
    What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:43










  • $begingroup$
    I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this
    $endgroup$
    – user607735
    Dec 7 '18 at 9:45






  • 1




    $begingroup$
    Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:47










  • $begingroup$
    Is the problem stated correctly, so one seat remains empty?
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:53










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 7 '18 at 10:07














  • 3




    $begingroup$
    What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:43










  • $begingroup$
    I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this
    $endgroup$
    – user607735
    Dec 7 '18 at 9:45






  • 1




    $begingroup$
    Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on?
    $endgroup$
    – Arthur
    Dec 7 '18 at 9:47










  • $begingroup$
    Is the problem stated correctly, so one seat remains empty?
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:53










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 7 '18 at 10:07








3




3




$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– Arthur
Dec 7 '18 at 9:43




$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– Arthur
Dec 7 '18 at 9:43












$begingroup$
I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this
$endgroup$
– user607735
Dec 7 '18 at 9:45




$begingroup$
I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this
$endgroup$
– user607735
Dec 7 '18 at 9:45




1




1




$begingroup$
Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on?
$endgroup$
– Arthur
Dec 7 '18 at 9:47




$begingroup$
Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on?
$endgroup$
– Arthur
Dec 7 '18 at 9:47












$begingroup$
Is the problem stated correctly, so one seat remains empty?
$endgroup$
– Christoph
Dec 7 '18 at 9:53




$begingroup$
Is the problem stated correctly, so one seat remains empty?
$endgroup$
– Christoph
Dec 7 '18 at 9:53












$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 10:07




$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 7 '18 at 10:07










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint: Consider Bob and Sally one unit taking two seats. So you have to place $n+1$ units at $n+2$ spots. In the end multiply by $2$ to account for Bob sitting on the left or right of Sally.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes so it could be writing as n+2 C n+1
    $endgroup$
    – user607735
    Dec 7 '18 at 9:55






  • 1




    $begingroup$
    Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:56



















0












$begingroup$

Treat Bob and Sally as the same person (say, V) that takes up $2$ spaces, i.e. you put it in chair $m$ and then you can't put anyone in chair $m+1$. This means V cannot be in chair $n+3$. So, we place $V$ first in $n+2$ seats, and since it takes up $2$ seats, we now have $n+1$ seats for $n$ people. This means that the number of permutations is $(n+2)!$. Now, we can have $2$ "states" for $V$: one where Bob sits on chair $m$ and Sally on chair $m+1$, or Sally sits on chair $m$ and Bob sits on $m+1$. So, we multiply by $2$ to account, for a grand total of $2(n+2)!$ seatings arrangements.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You can evaluate the arrangements separately.



    So for instance, the different arrangements in which Bob and Sally sit together are $$(n+2)times 2$$
    Note that deciding whether Bob or Sally sits to the right doubles the possibilities.



    Now, all possible arrangements in which $n$ different people sit (it doesn't matter whether together or not) is simply $$n!$$



    All possibilities are hence $$n!times (n+2)times2$$
    Note, however, that you also have to consider the gap; the total amount of possibilities is thus $$n!times (n+2)times 2times(n+1)=2times(n+2)!$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
      $endgroup$
      – Christoph
      Dec 7 '18 at 10:00












    • $begingroup$
      @Christoph you're definitely right!
      $endgroup$
      – Dr. Mathva
      Dec 7 '18 at 10:26











    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%2f3029701%2fhow-many-different-arrangements-are-there-if-bob-and-sally-must-always-be-seated%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









    2












    $begingroup$

    Hint: Consider Bob and Sally one unit taking two seats. So you have to place $n+1$ units at $n+2$ spots. In the end multiply by $2$ to account for Bob sitting on the left or right of Sally.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      yes so it could be writing as n+2 C n+1
      $endgroup$
      – user607735
      Dec 7 '18 at 9:55






    • 1




      $begingroup$
      Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
      $endgroup$
      – Christoph
      Dec 7 '18 at 9:56
















    2












    $begingroup$

    Hint: Consider Bob and Sally one unit taking two seats. So you have to place $n+1$ units at $n+2$ spots. In the end multiply by $2$ to account for Bob sitting on the left or right of Sally.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      yes so it could be writing as n+2 C n+1
      $endgroup$
      – user607735
      Dec 7 '18 at 9:55






    • 1




      $begingroup$
      Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
      $endgroup$
      – Christoph
      Dec 7 '18 at 9:56














    2












    2








    2





    $begingroup$

    Hint: Consider Bob and Sally one unit taking two seats. So you have to place $n+1$ units at $n+2$ spots. In the end multiply by $2$ to account for Bob sitting on the left or right of Sally.






    share|cite|improve this answer











    $endgroup$



    Hint: Consider Bob and Sally one unit taking two seats. So you have to place $n+1$ units at $n+2$ spots. In the end multiply by $2$ to account for Bob sitting on the left or right of Sally.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 10:01

























    answered Dec 7 '18 at 9:52









    ChristophChristoph

    12k1642




    12k1642












    • $begingroup$
      yes so it could be writing as n+2 C n+1
      $endgroup$
      – user607735
      Dec 7 '18 at 9:55






    • 1




      $begingroup$
      Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
      $endgroup$
      – Christoph
      Dec 7 '18 at 9:56


















    • $begingroup$
      yes so it could be writing as n+2 C n+1
      $endgroup$
      – user607735
      Dec 7 '18 at 9:55






    • 1




      $begingroup$
      Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
      $endgroup$
      – Christoph
      Dec 7 '18 at 9:56
















    $begingroup$
    yes so it could be writing as n+2 C n+1
    $endgroup$
    – user607735
    Dec 7 '18 at 9:55




    $begingroup$
    yes so it could be writing as n+2 C n+1
    $endgroup$
    – user607735
    Dec 7 '18 at 9:55




    1




    1




    $begingroup$
    Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:56




    $begingroup$
    Writing what as $binom{n+2}{n+1}$? The number of ways to seat the people is something else.
    $endgroup$
    – Christoph
    Dec 7 '18 at 9:56











    0












    $begingroup$

    Treat Bob and Sally as the same person (say, V) that takes up $2$ spaces, i.e. you put it in chair $m$ and then you can't put anyone in chair $m+1$. This means V cannot be in chair $n+3$. So, we place $V$ first in $n+2$ seats, and since it takes up $2$ seats, we now have $n+1$ seats for $n$ people. This means that the number of permutations is $(n+2)!$. Now, we can have $2$ "states" for $V$: one where Bob sits on chair $m$ and Sally on chair $m+1$, or Sally sits on chair $m$ and Bob sits on $m+1$. So, we multiply by $2$ to account, for a grand total of $2(n+2)!$ seatings arrangements.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Treat Bob and Sally as the same person (say, V) that takes up $2$ spaces, i.e. you put it in chair $m$ and then you can't put anyone in chair $m+1$. This means V cannot be in chair $n+3$. So, we place $V$ first in $n+2$ seats, and since it takes up $2$ seats, we now have $n+1$ seats for $n$ people. This means that the number of permutations is $(n+2)!$. Now, we can have $2$ "states" for $V$: one where Bob sits on chair $m$ and Sally on chair $m+1$, or Sally sits on chair $m$ and Bob sits on $m+1$. So, we multiply by $2$ to account, for a grand total of $2(n+2)!$ seatings arrangements.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Treat Bob and Sally as the same person (say, V) that takes up $2$ spaces, i.e. you put it in chair $m$ and then you can't put anyone in chair $m+1$. This means V cannot be in chair $n+3$. So, we place $V$ first in $n+2$ seats, and since it takes up $2$ seats, we now have $n+1$ seats for $n$ people. This means that the number of permutations is $(n+2)!$. Now, we can have $2$ "states" for $V$: one where Bob sits on chair $m$ and Sally on chair $m+1$, or Sally sits on chair $m$ and Bob sits on $m+1$. So, we multiply by $2$ to account, for a grand total of $2(n+2)!$ seatings arrangements.






        share|cite|improve this answer









        $endgroup$



        Treat Bob and Sally as the same person (say, V) that takes up $2$ spaces, i.e. you put it in chair $m$ and then you can't put anyone in chair $m+1$. This means V cannot be in chair $n+3$. So, we place $V$ first in $n+2$ seats, and since it takes up $2$ seats, we now have $n+1$ seats for $n$ people. This means that the number of permutations is $(n+2)!$. Now, we can have $2$ "states" for $V$: one where Bob sits on chair $m$ and Sally on chair $m+1$, or Sally sits on chair $m$ and Bob sits on $m+1$. So, we multiply by $2$ to account, for a grand total of $2(n+2)!$ seatings arrangements.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 9:56









        KykyKyky

        460213




        460213























            0












            $begingroup$

            You can evaluate the arrangements separately.



            So for instance, the different arrangements in which Bob and Sally sit together are $$(n+2)times 2$$
            Note that deciding whether Bob or Sally sits to the right doubles the possibilities.



            Now, all possible arrangements in which $n$ different people sit (it doesn't matter whether together or not) is simply $$n!$$



            All possibilities are hence $$n!times (n+2)times2$$
            Note, however, that you also have to consider the gap; the total amount of possibilities is thus $$n!times (n+2)times 2times(n+1)=2times(n+2)!$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
              $endgroup$
              – Christoph
              Dec 7 '18 at 10:00












            • $begingroup$
              @Christoph you're definitely right!
              $endgroup$
              – Dr. Mathva
              Dec 7 '18 at 10:26
















            0












            $begingroup$

            You can evaluate the arrangements separately.



            So for instance, the different arrangements in which Bob and Sally sit together are $$(n+2)times 2$$
            Note that deciding whether Bob or Sally sits to the right doubles the possibilities.



            Now, all possible arrangements in which $n$ different people sit (it doesn't matter whether together or not) is simply $$n!$$



            All possibilities are hence $$n!times (n+2)times2$$
            Note, however, that you also have to consider the gap; the total amount of possibilities is thus $$n!times (n+2)times 2times(n+1)=2times(n+2)!$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
              $endgroup$
              – Christoph
              Dec 7 '18 at 10:00












            • $begingroup$
              @Christoph you're definitely right!
              $endgroup$
              – Dr. Mathva
              Dec 7 '18 at 10:26














            0












            0








            0





            $begingroup$

            You can evaluate the arrangements separately.



            So for instance, the different arrangements in which Bob and Sally sit together are $$(n+2)times 2$$
            Note that deciding whether Bob or Sally sits to the right doubles the possibilities.



            Now, all possible arrangements in which $n$ different people sit (it doesn't matter whether together or not) is simply $$n!$$



            All possibilities are hence $$n!times (n+2)times2$$
            Note, however, that you also have to consider the gap; the total amount of possibilities is thus $$n!times (n+2)times 2times(n+1)=2times(n+2)!$$






            share|cite|improve this answer









            $endgroup$



            You can evaluate the arrangements separately.



            So for instance, the different arrangements in which Bob and Sally sit together are $$(n+2)times 2$$
            Note that deciding whether Bob or Sally sits to the right doubles the possibilities.



            Now, all possible arrangements in which $n$ different people sit (it doesn't matter whether together or not) is simply $$n!$$



            All possibilities are hence $$n!times (n+2)times2$$
            Note, however, that you also have to consider the gap; the total amount of possibilities is thus $$n!times (n+2)times 2times(n+1)=2times(n+2)!$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 7 '18 at 9:58









            Dr. MathvaDr. Mathva

            1,102317




            1,102317












            • $begingroup$
              You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
              $endgroup$
              – Christoph
              Dec 7 '18 at 10:00












            • $begingroup$
              @Christoph you're definitely right!
              $endgroup$
              – Dr. Mathva
              Dec 7 '18 at 10:26


















            • $begingroup$
              You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
              $endgroup$
              – Christoph
              Dec 7 '18 at 10:00












            • $begingroup$
              @Christoph you're definitely right!
              $endgroup$
              – Dr. Mathva
              Dec 7 '18 at 10:26
















            $begingroup$
            You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
            $endgroup$
            – Christoph
            Dec 7 '18 at 10:00






            $begingroup$
            You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately.
            $endgroup$
            – Christoph
            Dec 7 '18 at 10:00














            $begingroup$
            @Christoph you're definitely right!
            $endgroup$
            – Dr. Mathva
            Dec 7 '18 at 10:26




            $begingroup$
            @Christoph you're definitely right!
            $endgroup$
            – Dr. Mathva
            Dec 7 '18 at 10:26


















            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%2f3029701%2fhow-many-different-arrangements-are-there-if-bob-and-sally-must-always-be-seated%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more