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

43.9k93355




43.9k93355










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











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    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

    11.9k1542




    11.9k1542












    • $begingroup$
      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

        444213




        444213























            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

            957316




            957316












            • $begingroup$
              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


















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            Berounka

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            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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            Sphinx de Gizeh

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            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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