Combinatorics of sums

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$begingroup$


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:
$$beginalign
4 &= 1 + 1 + 1 + 1\
4 &= 2 + 1 + 1\
4 &= 1 + 2 + 1\
4 &= 1 + 1 + 2\
4 &= 2 + 2\
4 &= 3 + 1\
4 &= 1 + 3\
4 &= 4
endalign
$$

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?










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$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 6:25















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:
$$beginalign
4 &= 1 + 1 + 1 + 1\
4 &= 2 + 1 + 1\
4 &= 1 + 2 + 1\
4 &= 1 + 1 + 2\
4 &= 2 + 2\
4 &= 3 + 1\
4 &= 1 + 3\
4 &= 4
endalign
$$

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?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 6:25













2












2








2


0



$begingroup$


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:
$$beginalign
4 &= 1 + 1 + 1 + 1\
4 &= 2 + 1 + 1\
4 &= 1 + 2 + 1\
4 &= 1 + 1 + 2\
4 &= 2 + 2\
4 &= 3 + 1\
4 &= 1 + 3\
4 &= 4
endalign
$$

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?










share|cite|improve this question











$endgroup$




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:
$$beginalign
4 &= 1 + 1 + 1 + 1\
4 &= 2 + 1 + 1\
4 &= 1 + 2 + 1\
4 &= 1 + 1 + 2\
4 &= 2 + 2\
4 &= 3 + 1\
4 &= 1 + 3\
4 &= 4
endalign
$$

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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edited Jan 20 at 3:05









user549397

1,5061418




1,5061418










asked Jan 20 at 1:36









shadoxshadox

1133




1133







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 6:25












  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_(combinatorics)
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 6:25







1




1




$begingroup$
en.wikipedia.org/wiki/Composition_(combinatorics)
$endgroup$
– Lord Shark the Unknown
Jan 20 at 6:25




$begingroup$
en.wikipedia.org/wiki/Composition_(combinatorics)
$endgroup$
– Lord Shark the Unknown
Jan 20 at 6:25










2 Answers
2






active

oldest

votes


















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




    $begingroup$
    Oh, I should've thought about that! It seems to be closest to a theory I need.
    $endgroup$
    – shadox
    Jan 20 at 12:56


















7












$begingroup$

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









$endgroup$












  • $begingroup$
    I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
    $endgroup$
    – shadox
    Jan 20 at 13:00










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









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




    $begingroup$
    Oh, I should've thought about that! It seems to be closest to a theory I need.
    $endgroup$
    – shadox
    Jan 20 at 12:56















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




    $begingroup$
    Oh, I should've thought about that! It seems to be closest to a theory I need.
    $endgroup$
    – shadox
    Jan 20 at 12:56













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 Jan 20 at 1:49









Michael WangMichael Wang

171114




171114







  • 1




    $begingroup$
    Oh, I should've thought about that! It seems to be closest to a theory I need.
    $endgroup$
    – shadox
    Jan 20 at 12:56












  • 1




    $begingroup$
    Oh, I should've thought about that! It seems to be closest to a theory I need.
    $endgroup$
    – shadox
    Jan 20 at 12:56







1




1




$begingroup$
Oh, I should've thought about that! It seems to be closest to a theory I need.
$endgroup$
– shadox
Jan 20 at 12:56




$begingroup$
Oh, I should've thought about that! It seems to be closest to a theory I need.
$endgroup$
– shadox
Jan 20 at 12:56











7












$begingroup$

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









$endgroup$












  • $begingroup$
    I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
    $endgroup$
    – shadox
    Jan 20 at 13:00















7












$begingroup$

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









$endgroup$












  • $begingroup$
    I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
    $endgroup$
    – shadox
    Jan 20 at 13:00













7












7








7





$begingroup$

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 Jan 20 at 1:40









angryavianangryavian

40.7k23380




40.7k23380











  • $begingroup$
    I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
    $endgroup$
    – shadox
    Jan 20 at 13:00
















  • $begingroup$
    I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
    $endgroup$
    – shadox
    Jan 20 at 13:00















$begingroup$
I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
$endgroup$
– shadox
Jan 20 at 13:00




$begingroup$
I figured this out but wasn't really aware if there is a theory in place that explains that or I have to prove through mathematical induction that my formula stands for any number S.
$endgroup$
– shadox
Jan 20 at 13:00

















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