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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  • $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    2 mins ago
















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.







$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    2 mins ago














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









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edited 3 hours ago









Philip

1,1941315




1,1941315






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asked 4 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.
Check out our Code of Conduct.






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












  • $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    2 mins ago


















  • $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    2 mins ago
















$begingroup$
en.wikipedia.org/wiki/Composition_(combinatorics)
$endgroup$
– Lord Shark the Unknown
2 mins ago




$begingroup$
en.wikipedia.org/wiki/Composition_(combinatorics)
$endgroup$
– Lord Shark the Unknown
2 mins ago










2 Answers
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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









$endgroup$





















    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 $2^{S-1}$.






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      2 Answers
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      2 Answers
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      5












      $begingroup$

      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









      $endgroup$


















        5












        $begingroup$

        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









        $endgroup$
















          5












          5








          5





          $begingroup$

          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 4 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 4 hours ago









                  Michael WangMichael Wang

                  11210




                  11210






















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