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 42 mins ago
Philip
1,1941315
asked 2 hours ago
shadoxshadox
1112
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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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$endgroup$
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 42 mins ago
Philip
1,1941315
asked 2 hours ago
shadoxshadox
1112
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.
$endgroup$
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 42 mins ago
Philip
1,1941315
asked 2 hours ago
shadoxshadox
1112
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
edited 42 mins ago
Philip
1,1941315
asked 2 hours ago
shadoxshadox
1112
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.
edited 42 mins ago
Philip
1,1941315
asked 2 hours ago
shadoxshadox
1112
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.
edited 42 mins ago
Philip
1,1941315
edited 42 mins ago
Philip
1,1941315
edited 42 mins ago
Philip
1,1941315
1,1941315
asked 2 hours ago
shadoxshadox
1112
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.
asked 2 hours ago
shadoxshadox
1112
asked 2 hours ago
shadoxshadox
1112
1112
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.
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.
add a comment |
add a comment |
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.
answered 2 hours ago
angryavianangryavian
40.2k23280
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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 1 hour ago
Michael WangMichael Wang
10210
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
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 2 hours ago
angryavianangryavian
40.2k23280
$endgroup$
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 2 hours ago
angryavianangryavian
40.2k23280
$endgroup$
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 2 hours ago
angryavianangryavian
40.2k23280
$endgroup$
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 2 hours ago
angryavianangryavian
40.2k23280
answered 2 hours ago
angryavianangryavian
40.2k23280
answered 2 hours ago
angryavianangryavian
40.2k23280
answered 2 hours ago
angryavianangryavian
40.2k23280
40.2k23280
add a comment |
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 1 hour ago
Michael WangMichael Wang
10210
$endgroup$
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 1 hour ago
Michael WangMichael Wang
10210
$endgroup$
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 1 hour ago
Michael WangMichael Wang
10210
$endgroup$
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 1 hour ago
Michael WangMichael Wang
10210
answered 1 hour ago
Michael WangMichael Wang
10210
answered 1 hour ago
Michael WangMichael Wang
10210
answered 1 hour ago
Michael WangMichael Wang
10210
10210
add a comment |
add a comment |
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