Combinatorics of sums
$begingroup$
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
edited Jan 20 at 3:05
user549397
1,5101418
asked Jan 20 at 1:36
shadoxshadox
1133
$endgroup$
add a comment |
$begingroup$
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
edited Jan 20 at 3:05
user549397
1,5101418
asked Jan 20 at 1:36
shadoxshadox
1133
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Composition_(combinatorics)
$endgroup$
– Lord Shark the Unknown
Jan 20 at 6:25
add a comment |
$begingroup$
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
edited Jan 20 at 3:05
user549397
1,5101418
asked Jan 20 at 1:36
shadoxshadox
1133
$endgroup$
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
edited Jan 20 at 3:05
user549397
1,5101418
asked Jan 20 at 1:36
shadoxshadox
1133
edited Jan 20 at 3:05
user549397
1,5101418
asked Jan 20 at 1:36
shadoxshadox
1133
edited Jan 20 at 3:05
user549397
1,5101418
edited Jan 20 at 3:05
user549397
1,5101418
edited Jan 20 at 3:05
user549397
1,5101418
1,5101418
asked Jan 20 at 1:36
shadoxshadox
1133
asked Jan 20 at 1:36
shadoxshadox
1133
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $2^{S-1}$.
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $2^{S-1}$.
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
$endgroup$
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
add a comment |
$begingroup$
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 $2^{S-1}$.
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
$endgroup$
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
add a comment |
$begingroup$
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 $2^{S-1}$.
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
$endgroup$
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 $2^{S-1}$.
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
answered Jan 20 at 1:49
Michael WangMichael Wang
186215
186215
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
add a comment |
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
add a comment |
$begingroup$
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.
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
$endgroup$
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.
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
answered Jan 20 at 1:40
angryavianangryavian
41.3k23380
41.3k23380
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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1
$begingroup$
en.wikipedia.org/wiki/Composition_(combinatorics)
$endgroup$
– Lord Shark the Unknown
Jan 20 at 6:25