How many ways to color each cell of $2times 13$ table with $4$ colors so that cells with a common side aren't...












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$2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.



How many ways to do this are there?



Thanks in a lot advance!






combinatorics







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    0












    $begingroup$


    $2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.



    How many ways to do this are there?



    Thanks in a lot advance!






    combinatorics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      $2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.



      How many ways to do this are there?



      Thanks in a lot advance!






      combinatorics







      share|cite|improve this question









      $endgroup$




      $2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.



      How many ways to do this are there?



      Thanks in a lot advance!






      combinatorics




      combinatorics






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 20 at 14:24









      Xxx DddXxx Ddd

      354




      354






















          3 Answers
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          $begingroup$

          Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.






          share|cite|improve this answer









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            2












            $begingroup$

            There are $4times3=12$ possibilities for the first column.



            If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.



            This leads to $12times7^{12}$ possibilities.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.



              If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.



              enter image description here



              Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).



              Then we proceed as follows:




              • First Column in 12 ways:


                • Choose 1 in 4 ways.

                • Choose 14 in 3 ways.



              • Adjacent Columns in $3^{12} times 5^{12}$ ways:


                • Choose 2 to 13 in $3^{12}$ ways.

                • Choose 15 to 26 in $5^{12}$ ways:


                  • 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,

                  • 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.






              So we can choose in $12 times 3^{12} times {5}^{12}$ ways.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                $endgroup$
                – Ross Millikan
                Jan 20 at 15:12










              • $begingroup$
                I stated 4 colors in the post, thanks anyway
                $endgroup$
                – Xxx Ddd
                Jan 20 at 15:13










              • $begingroup$
                @Ross right... I hadn't considered that.
                $endgroup$
                – Sagnik
                Jan 20 at 15:18











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              3 Answers
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              3 Answers
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              2












              $begingroup$

              Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.






                  share|cite|improve this answer









                  $endgroup$



                  Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 20 at 14:50









                  Christian BlatterChristian Blatter

                  174k8115327




                  174k8115327























                      2












                      $begingroup$

                      There are $4times3=12$ possibilities for the first column.



                      If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.



                      This leads to $12times7^{12}$ possibilities.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        There are $4times3=12$ possibilities for the first column.



                        If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.



                        This leads to $12times7^{12}$ possibilities.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          There are $4times3=12$ possibilities for the first column.



                          If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.



                          This leads to $12times7^{12}$ possibilities.






                          share|cite|improve this answer









                          $endgroup$



                          There are $4times3=12$ possibilities for the first column.



                          If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.



                          This leads to $12times7^{12}$ possibilities.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 20 at 14:51









                          drhabdrhab

                          102k545136




                          102k545136























                              0












                              $begingroup$

                              A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.



                              If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.



                              enter image description here



                              Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).



                              Then we proceed as follows:




                              • First Column in 12 ways:


                                • Choose 1 in 4 ways.

                                • Choose 14 in 3 ways.



                              • Adjacent Columns in $3^{12} times 5^{12}$ ways:


                                • Choose 2 to 13 in $3^{12}$ ways.

                                • Choose 15 to 26 in $5^{12}$ ways:


                                  • 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,

                                  • 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.






                              So we can choose in $12 times 3^{12} times {5}^{12}$ ways.






                              share|cite|improve this answer











                              $endgroup$









                              • 1




                                $begingroup$
                                You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                                $endgroup$
                                – Ross Millikan
                                Jan 20 at 15:12










                              • $begingroup$
                                I stated 4 colors in the post, thanks anyway
                                $endgroup$
                                – Xxx Ddd
                                Jan 20 at 15:13










                              • $begingroup$
                                @Ross right... I hadn't considered that.
                                $endgroup$
                                – Sagnik
                                Jan 20 at 15:18
















                              0












                              $begingroup$

                              A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.



                              If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.



                              enter image description here



                              Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).



                              Then we proceed as follows:




                              • First Column in 12 ways:


                                • Choose 1 in 4 ways.

                                • Choose 14 in 3 ways.



                              • Adjacent Columns in $3^{12} times 5^{12}$ ways:


                                • Choose 2 to 13 in $3^{12}$ ways.

                                • Choose 15 to 26 in $5^{12}$ ways:


                                  • 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,

                                  • 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.






                              So we can choose in $12 times 3^{12} times {5}^{12}$ ways.






                              share|cite|improve this answer











                              $endgroup$









                              • 1




                                $begingroup$
                                You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                                $endgroup$
                                – Ross Millikan
                                Jan 20 at 15:12










                              • $begingroup$
                                I stated 4 colors in the post, thanks anyway
                                $endgroup$
                                – Xxx Ddd
                                Jan 20 at 15:13










                              • $begingroup$
                                @Ross right... I hadn't considered that.
                                $endgroup$
                                – Sagnik
                                Jan 20 at 15:18














                              0












                              0








                              0





                              $begingroup$

                              A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.



                              If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.



                              enter image description here



                              Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).



                              Then we proceed as follows:




                              • First Column in 12 ways:


                                • Choose 1 in 4 ways.

                                • Choose 14 in 3 ways.



                              • Adjacent Columns in $3^{12} times 5^{12}$ ways:


                                • Choose 2 to 13 in $3^{12}$ ways.

                                • Choose 15 to 26 in $5^{12}$ ways:


                                  • 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,

                                  • 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.






                              So we can choose in $12 times 3^{12} times {5}^{12}$ ways.






                              share|cite|improve this answer











                              $endgroup$



                              A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.



                              If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.



                              enter image description here



                              Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).



                              Then we proceed as follows:




                              • First Column in 12 ways:


                                • Choose 1 in 4 ways.

                                • Choose 14 in 3 ways.



                              • Adjacent Columns in $3^{12} times 5^{12}$ ways:


                                • Choose 2 to 13 in $3^{12}$ ways.

                                • Choose 15 to 26 in $5^{12}$ ways:


                                  • 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,

                                  • 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.






                              So we can choose in $12 times 3^{12} times {5}^{12}$ ways.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 20 at 15:55

























                              answered Jan 20 at 15:08









                              SagnikSagnik

                              1377




                              1377








                              • 1




                                $begingroup$
                                You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                                $endgroup$
                                – Ross Millikan
                                Jan 20 at 15:12










                              • $begingroup$
                                I stated 4 colors in the post, thanks anyway
                                $endgroup$
                                – Xxx Ddd
                                Jan 20 at 15:13










                              • $begingroup$
                                @Ross right... I hadn't considered that.
                                $endgroup$
                                – Sagnik
                                Jan 20 at 15:18














                              • 1




                                $begingroup$
                                You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                                $endgroup$
                                – Ross Millikan
                                Jan 20 at 15:12










                              • $begingroup$
                                I stated 4 colors in the post, thanks anyway
                                $endgroup$
                                – Xxx Ddd
                                Jan 20 at 15:13










                              • $begingroup$
                                @Ross right... I hadn't considered that.
                                $endgroup$
                                – Sagnik
                                Jan 20 at 15:18








                              1




                              1




                              $begingroup$
                              You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                              $endgroup$
                              – Ross Millikan
                              Jan 20 at 15:12




                              $begingroup$
                              You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
                              $endgroup$
                              – Ross Millikan
                              Jan 20 at 15:12












                              $begingroup$
                              I stated 4 colors in the post, thanks anyway
                              $endgroup$
                              – Xxx Ddd
                              Jan 20 at 15:13




                              $begingroup$
                              I stated 4 colors in the post, thanks anyway
                              $endgroup$
                              – Xxx Ddd
                              Jan 20 at 15:13












                              $begingroup$
                              @Ross right... I hadn't considered that.
                              $endgroup$
                              – Sagnik
                              Jan 20 at 15:18




                              $begingroup$
                              @Ross right... I hadn't considered that.
                              $endgroup$
                              – Sagnik
                              Jan 20 at 15:18


















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