Arranging cats and dogs - what is wrong with my approach












7












$begingroup$


We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?



Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.



I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)



The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.



The cats can now be arranged in $3!$ ways.



So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$



Where have I gone wrong?






combinatorics permutations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Is your problem arising as a consequence of a rainfall ?
    $endgroup$
    – Jean Marie
    Mar 30 at 16:27
















7












$begingroup$


We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?



Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.



I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)



The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.



The cats can now be arranged in $3!$ ways.



So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$



Where have I gone wrong?






combinatorics permutations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Is your problem arising as a consequence of a rainfall ?
    $endgroup$
    – Jean Marie
    Mar 30 at 16:27














7












7








7


1



$begingroup$


We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?



Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.



I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)



The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.



The cats can now be arranged in $3!$ ways.



So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$



Where have I gone wrong?






combinatorics permutations







share|cite|improve this question









$endgroup$




We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?



Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.



I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)



The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.



The cats can now be arranged in $3!$ ways.



So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$



Where have I gone wrong?






combinatorics permutations




combinatorics permutations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 30 at 11:26









Akshat AgarwalAkshat Agarwal

513




513








  • 2




    $begingroup$
    Is your problem arising as a consequence of a rainfall ?
    $endgroup$
    – Jean Marie
    Mar 30 at 16:27














  • 2




    $begingroup$
    Is your problem arising as a consequence of a rainfall ?
    $endgroup$
    – Jean Marie
    Mar 30 at 16:27








2




2




$begingroup$
Is your problem arising as a consequence of a rainfall ?
$endgroup$
– Jean Marie
Mar 30 at 16:27




$begingroup$
Is your problem arising as a consequence of a rainfall ?
$endgroup$
– Jean Marie
Mar 30 at 16:27










2 Answers
2






active

oldest

votes


















6












$begingroup$

The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$



You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:



$$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$



So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?



Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.



Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You can separate the two cases for the two last dogs: single dogs and double dogs.



    Single dogs:
    $$P(4,2)=frac{4!}{2!}=12.$$
    Double dogs:
    $$P(2,2)cdot C(4,1)=2cdot 4=8.$$
    Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.



    The final answer is:
    $$3!cdot 20cdot 4cdot 3=1440.$$






    share|cite|improve this answer









    $endgroup$














      Your Answer








      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%2f3168195%2farranging-cats-and-dogs-what-is-wrong-with-my-approach%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$



      You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:



      $$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$



      So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?



      Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.



      Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$



        You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:



        $$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$



        So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?



        Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.



        Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$



          You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:



          $$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$



          So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?



          Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.



          Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$






          share|cite|improve this answer











          $endgroup$



          The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$



          You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:



          $$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$



          So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?



          Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.



          Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 30 at 15:26

























          answered Mar 30 at 11:44









          lulululu

          43.6k25081




          43.6k25081























              0












              $begingroup$

              You can separate the two cases for the two last dogs: single dogs and double dogs.



              Single dogs:
              $$P(4,2)=frac{4!}{2!}=12.$$
              Double dogs:
              $$P(2,2)cdot C(4,1)=2cdot 4=8.$$
              Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.



              The final answer is:
              $$3!cdot 20cdot 4cdot 3=1440.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You can separate the two cases for the two last dogs: single dogs and double dogs.



                Single dogs:
                $$P(4,2)=frac{4!}{2!}=12.$$
                Double dogs:
                $$P(2,2)cdot C(4,1)=2cdot 4=8.$$
                Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.



                The final answer is:
                $$3!cdot 20cdot 4cdot 3=1440.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You can separate the two cases for the two last dogs: single dogs and double dogs.



                  Single dogs:
                  $$P(4,2)=frac{4!}{2!}=12.$$
                  Double dogs:
                  $$P(2,2)cdot C(4,1)=2cdot 4=8.$$
                  Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.



                  The final answer is:
                  $$3!cdot 20cdot 4cdot 3=1440.$$






                  share|cite|improve this answer









                  $endgroup$



                  You can separate the two cases for the two last dogs: single dogs and double dogs.



                  Single dogs:
                  $$P(4,2)=frac{4!}{2!}=12.$$
                  Double dogs:
                  $$P(2,2)cdot C(4,1)=2cdot 4=8.$$
                  Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.



                  The final answer is:
                  $$3!cdot 20cdot 4cdot 3=1440.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 30 at 15:27









                  farruhotafarruhota

                  22k2942




                  22k2942






























                      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%2f3168195%2farranging-cats-and-dogs-what-is-wrong-with-my-approach%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

                      If I really need a card on my start hand, how many mulligans make sense? [duplicate]

                      Image
                      5 This question already has an answer here: How do you calculate the likelihood of drawing certain cards in your opening hand? 3 answers There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt? magic-the-gathering probability share | improve this question asked Nov 23 '18 at 12:12 dan dan 152 4
                      Read more

                      Alcedinidae

                      Image
                      Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
                      Read more

                      Can an atomic nucleus contain both particles and antiparticles? [duplicate]

                      Image
                      9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
                      Read more