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









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

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      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






























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