Combinatorics of sums












2












$begingroup$


Let's say that we have a number S that represents a sum. This sum can be broken down into a sum of terms. I want to calculate how many expressions I can write that represent that sum where terms are in the range from $ 1 $ to $ S $.




Example:
$$begin{align}
4 &= 1 + 1 + 1 + 1\
4 &= 2 + 1 + 1\
4 &= 1 + 2 + 1\
4 &= 1 + 1 + 2\
4 &= 2 + 2\
4 &= 3 + 1\
4 &= 1 + 3\
4 &= 4
end{align}
$$
For $S=4$ we have $N=8$ .

For $S=3$, we have $N=4$




Although I figured out that I can calculate that with this formula:



$N = 2^{S-1}$



I can't really tell why is that. I can count them for few sums and see the rule but is there a better way to explain this?






combinatorics combinations







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shadox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2












    $begingroup$


    Let's say that we have a number S that represents a sum. This sum can be broken down into a sum of terms. I want to calculate how many expressions I can write that represent that sum where terms are in the range from $ 1 $ to $ S $.




    Example:
    $$begin{align}
    4 &= 1 + 1 + 1 + 1\
    4 &= 2 + 1 + 1\
    4 &= 1 + 2 + 1\
    4 &= 1 + 1 + 2\
    4 &= 2 + 2\
    4 &= 3 + 1\
    4 &= 1 + 3\
    4 &= 4
    end{align}
    $$
    For $S=4$ we have $N=8$ .

    For $S=3$, we have $N=4$




    Although I figured out that I can calculate that with this formula:



    $N = 2^{S-1}$



    I can't really tell why is that. I can count them for few sums and see the rule but is there a better way to explain this?






    combinatorics combinations







    share|cite|improve this question









    New contributor




    shadox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      Let's say that we have a number S that represents a sum. This sum can be broken down into a sum of terms. I want to calculate how many expressions I can write that represent that sum where terms are in the range from $ 1 $ to $ S $.




      Example:
      $$begin{align}
      4 &= 1 + 1 + 1 + 1\
      4 &= 2 + 1 + 1\
      4 &= 1 + 2 + 1\
      4 &= 1 + 1 + 2\
      4 &= 2 + 2\
      4 &= 3 + 1\
      4 &= 1 + 3\
      4 &= 4
      end{align}
      $$
      For $S=4$ we have $N=8$ .

      For $S=3$, we have $N=4$




      Although I figured out that I can calculate that with this formula:



      $N = 2^{S-1}$



      I can't really tell why is that. I can count them for few sums and see the rule but is there a better way to explain this?






      combinatorics combinations







      share|cite|improve this question









      New contributor




      shadox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let's say that we have a number S that represents a sum. This sum can be broken down into a sum of terms. I want to calculate how many expressions I can write that represent that sum where terms are in the range from $ 1 $ to $ S $.




      Example:
      $$begin{align}
      4 &= 1 + 1 + 1 + 1\
      4 &= 2 + 1 + 1\
      4 &= 1 + 2 + 1\
      4 &= 1 + 1 + 2\
      4 &= 2 + 2\
      4 &= 3 + 1\
      4 &= 1 + 3\
      4 &= 4
      end{align}
      $$
      For $S=4$ we have $N=8$ .

      For $S=3$, we have $N=4$




      Although I figured out that I can calculate that with this formula:



      $N = 2^{S-1}$



      I can't really tell why is that. I can count them for few sums and see the rule but is there a better way to explain this?






      combinatorics combinations




      combinatorics combinations






      share|cite|improve this question









      New contributor




      shadox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









      New contributor




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      Check out our Code of Conduct.









      share|cite|improve this question




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      edited 47 mins ago









      Philip

      1,1941315




      1,1941315






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      asked 2 hours ago









      shadoxshadox

      1112




      1112




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      New contributor





      shadox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          $begingroup$

          Hint: Imagine $S$ balls lined up in a row. Between each pair of adjacent balls, you can choose whether to put a "divider" between them or not. After considering all such possible pairs, you can group the balls according to the dividers. For example, $4=1+1+2$ can be depicted as *|*|** where | denotes a divider, and * denotes a ball.






          share|cite|improve this answer









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            1












            $begingroup$

            See Pascal's triangle. Notice how each subsequent line is created by summing up its previous lines. For example, let's see the 4 example. See the fourth line, which says 1 3 3 1. This means there are 1 way to sum 4 with 4 integers, 3 ways to sum 4 with 3 integers, 3 ways to sum 4 with 2 integers, and 1 way to sum 4 into 1 integer. And the sum of each row of Pascal's triangle is $2^{S-1}$.






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              2 Answers
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              2 Answers
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              active

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              $begingroup$

              Hint: Imagine $S$ balls lined up in a row. Between each pair of adjacent balls, you can choose whether to put a "divider" between them or not. After considering all such possible pairs, you can group the balls according to the dividers. For example, $4=1+1+2$ can be depicted as *|*|** where | denotes a divider, and * denotes a ball.






              share|cite|improve this answer









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                3












                $begingroup$

                Hint: Imagine $S$ balls lined up in a row. Between each pair of adjacent balls, you can choose whether to put a "divider" between them or not. After considering all such possible pairs, you can group the balls according to the dividers. For example, $4=1+1+2$ can be depicted as *|*|** where | denotes a divider, and * denotes a ball.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Hint: Imagine $S$ balls lined up in a row. Between each pair of adjacent balls, you can choose whether to put a "divider" between them or not. After considering all such possible pairs, you can group the balls according to the dividers. For example, $4=1+1+2$ can be depicted as *|*|** where | denotes a divider, and * denotes a ball.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Imagine $S$ balls lined up in a row. Between each pair of adjacent balls, you can choose whether to put a "divider" between them or not. After considering all such possible pairs, you can group the balls according to the dividers. For example, $4=1+1+2$ can be depicted as *|*|** where | denotes a divider, and * denotes a ball.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  angryavianangryavian

                  40.2k23280




                  40.2k23280























                      1












                      $begingroup$

                      See Pascal's triangle. Notice how each subsequent line is created by summing up its previous lines. For example, let's see the 4 example. See the fourth line, which says 1 3 3 1. This means there are 1 way to sum 4 with 4 integers, 3 ways to sum 4 with 3 integers, 3 ways to sum 4 with 2 integers, and 1 way to sum 4 into 1 integer. And the sum of each row of Pascal's triangle is $2^{S-1}$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        See Pascal's triangle. Notice how each subsequent line is created by summing up its previous lines. For example, let's see the 4 example. See the fourth line, which says 1 3 3 1. This means there are 1 way to sum 4 with 4 integers, 3 ways to sum 4 with 3 integers, 3 ways to sum 4 with 2 integers, and 1 way to sum 4 into 1 integer. And the sum of each row of Pascal's triangle is $2^{S-1}$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          See Pascal's triangle. Notice how each subsequent line is created by summing up its previous lines. For example, let's see the 4 example. See the fourth line, which says 1 3 3 1. This means there are 1 way to sum 4 with 4 integers, 3 ways to sum 4 with 3 integers, 3 ways to sum 4 with 2 integers, and 1 way to sum 4 into 1 integer. And the sum of each row of Pascal's triangle is $2^{S-1}$.






                          share|cite|improve this answer









                          $endgroup$



                          See Pascal's triangle. Notice how each subsequent line is created by summing up its previous lines. For example, let's see the 4 example. See the fourth line, which says 1 3 3 1. This means there are 1 way to sum 4 with 4 integers, 3 ways to sum 4 with 3 integers, 3 ways to sum 4 with 2 integers, and 1 way to sum 4 into 1 integer. And the sum of each row of Pascal's triangle is $2^{S-1}$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          Michael WangMichael Wang

                          10210




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