Mixing
Sum of each combination
$begingroup$
A notation question.
Let's say I have a group of $N$ elements. I want to pick $n$ elements within that group and make the following computation (let's assume $N=3$, $n=1$, and the elements are $x_i$, where $i$ is between $0$ and $N$):
$$ x_1cdot(1-x_2)cdot(1-x_3)+x_2cdot (1-x_1)cdot (1-x_3)+x_3cdot(1-x_1)cdot(1-x_2).$$
As you probably understand, when $n=1$ I pick one element from the group and multiply it with the "one minus" of all the rest in the group (which weren't picked). I do so for each combination.
Another example, for $n=2$, I would do:
$$x_1cdot x_2 cdot (1-x_3) + x_1cdot x_3 cdot (1-x_2) + x_2cdot x_3 cdot (1-x_1).$$
I want to formulate it for a general $N$ and $n$.
Obviously, there suppose to be a $sum$, and then a multiplication between two groups of $prod$, something like:
$$ sum_{all-combinations}left[prod_{picked-elements}x_icdot prod_{non-picked-elements}(1-x_j)right].$$
The thing is that I do not know how to write the groups properly.
Can someone please advise? Thanks!
combinatorics combinations
asked Jan 4 at 23:13
GilshoGilsho
33
$endgroup$
add a comment |
$begingroup$
A notation question.
Let's say I have a group of $N$ elements. I want to pick $n$ elements within that group and make the following computation (let's assume $N=3$, $n=1$, and the elements are $x_i$, where $i$ is between $0$ and $N$):
$$ x_1cdot(1-x_2)cdot(1-x_3)+x_2cdot (1-x_1)cdot (1-x_3)+x_3cdot(1-x_1)cdot(1-x_2).$$
As you probably understand, when $n=1$ I pick one element from the group and multiply it with the "one minus" of all the rest in the group (which weren't picked). I do so for each combination.
Another example, for $n=2$, I would do:
$$x_1cdot x_2 cdot (1-x_3) + x_1cdot x_3 cdot (1-x_2) + x_2cdot x_3 cdot (1-x_1).$$
I want to formulate it for a general $N$ and $n$.
Obviously, there suppose to be a $sum$, and then a multiplication between two groups of $prod$, something like:
$$ sum_{all-combinations}left[prod_{picked-elements}x_icdot prod_{non-picked-elements}(1-x_j)right].$$
The thing is that I do not know how to write the groups properly.
Can someone please advise? Thanks!
combinatorics combinations
asked Jan 4 at 23:13
GilshoGilsho
33
$endgroup$
$begingroup$
Let $X$ be your group of $N$ elements. Define $P$ (short for pairs) to be the set of ordered pairs $P = {(A,B)~:~Acap B = emptyset,~Acup B = X,~|A|=n}$. Your sum can then be written as $sumlimits_{(A,B)in P}left(prodlimits_{xin A}x~cdot~prodlimits_{xin B}(1-x)right)$. There are surely other ways to write this, but this is just one of the first to come to my mind.
$endgroup$
– JMoravitz
Jan 4 at 23:21
add a comment |
$begingroup$
A notation question.
Let's say I have a group of $N$ elements. I want to pick $n$ elements within that group and make the following computation (let's assume $N=3$, $n=1$, and the elements are $x_i$, where $i$ is between $0$ and $N$):
$$ x_1cdot(1-x_2)cdot(1-x_3)+x_2cdot (1-x_1)cdot (1-x_3)+x_3cdot(1-x_1)cdot(1-x_2).$$
As you probably understand, when $n=1$ I pick one element from the group and multiply it with the "one minus" of all the rest in the group (which weren't picked). I do so for each combination.
Another example, for $n=2$, I would do:
$$x_1cdot x_2 cdot (1-x_3) + x_1cdot x_3 cdot (1-x_2) + x_2cdot x_3 cdot (1-x_1).$$
I want to formulate it for a general $N$ and $n$.
Obviously, there suppose to be a $sum$, and then a multiplication between two groups of $prod$, something like:
$$ sum_{all-combinations}left[prod_{picked-elements}x_icdot prod_{non-picked-elements}(1-x_j)right].$$
The thing is that I do not know how to write the groups properly.
Can someone please advise? Thanks!
combinatorics combinations
asked Jan 4 at 23:13
GilshoGilsho
33
$endgroup$
A notation question.
Let's say I have a group of $N$ elements. I want to pick $n$ elements within that group and make the following computation (let's assume $N=3$, $n=1$, and the elements are $x_i$, where $i$ is between $0$ and $N$):
$$ x_1cdot(1-x_2)cdot(1-x_3)+x_2cdot (1-x_1)cdot (1-x_3)+x_3cdot(1-x_1)cdot(1-x_2).$$
As you probably understand, when $n=1$ I pick one element from the group and multiply it with the "one minus" of all the rest in the group (which weren't picked). I do so for each combination.
Another example, for $n=2$, I would do:
$$x_1cdot x_2 cdot (1-x_3) + x_1cdot x_3 cdot (1-x_2) + x_2cdot x_3 cdot (1-x_1).$$
I want to formulate it for a general $N$ and $n$.
Obviously, there suppose to be a $sum$, and then a multiplication between two groups of $prod$, something like:
$$ sum_{all-combinations}left[prod_{picked-elements}x_icdot prod_{non-picked-elements}(1-x_j)right].$$
The thing is that I do not know how to write the groups properly.
Can someone please advise? Thanks!
combinatorics combinations
combinatorics combinations
asked Jan 4 at 23:13
GilshoGilsho
33
asked Jan 4 at 23:13
GilshoGilsho
33
asked Jan 4 at 23:13
GilshoGilsho
33
asked Jan 4 at 23:13
GilshoGilsho
33
asked Jan 4 at 23:13
GilshoGilsho
33
33
$begingroup$
Let $X$ be your group of $N$ elements. Define $P$ (short for pairs) to be the set of ordered pairs $P = {(A,B)~:~Acap B = emptyset,~Acup B = X,~|A|=n}$. Your sum can then be written as $sumlimits_{(A,B)in P}left(prodlimits_{xin A}x~cdot~prodlimits_{xin B}(1-x)right)$. There are surely other ways to write this, but this is just one of the first to come to my mind.
$endgroup$
– JMoravitz
Jan 4 at 23:21
add a comment |
$begingroup$
Let $X$ be your group of $N$ elements. Define $P$ (short for pairs) to be the set of ordered pairs $P = {(A,B)~:~Acap B = emptyset,~Acup B = X,~|A|=n}$. Your sum can then be written as $sumlimits_{(A,B)in P}left(prodlimits_{xin A}x~cdot~prodlimits_{xin B}(1-x)right)$. There are surely other ways to write this, but this is just one of the first to come to my mind.
$endgroup$
– JMoravitz
Jan 4 at 23:21
$begingroup$
Let $X$ be your group of $N$ elements. Define $P$ (short for pairs) to be the set of ordered pairs $P = {(A,B)~:~Acap B = emptyset,~Acup B = X,~|A|=n}$. Your sum can then be written as $sumlimits_{(A,B)in P}left(prodlimits_{xin A}x~cdot~prodlimits_{xin B}(1-x)right)$. There are surely other ways to write this, but this is just one of the first to come to my mind.
$endgroup$
– JMoravitz
Jan 4 at 23:21
$begingroup$
Let $X$ be your group of $N$ elements. Define $P$ (short for pairs) to be the set of ordered pairs $P = {(A,B)~:~Acap B = emptyset,~Acup B = X,~|A|=n}$. Your sum can then be written as $sumlimits_{(A,B)in P}left(prodlimits_{xin A}x~cdot~prodlimits_{xin B}(1-x)right)$. There are surely other ways to write this, but this is just one of the first to come to my mind.
$endgroup$
– JMoravitz
Jan 4 at 23:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the cleanest way is$$sum_{substack Asubseteq [N]\|A|=n}Bigg(prod_{iin A}x_iBigg)Bigg(prod_{iin [n]setminus A}(1-x_i)Bigg),$$
where $[N]={1,2,dots,N}$.
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
$endgroup$
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
I think the cleanest way is$$sum_{substack Asubseteq [N]\|A|=n}Bigg(prod_{iin A}x_iBigg)Bigg(prod_{iin [n]setminus A}(1-x_i)Bigg),$$
where $[N]={1,2,dots,N}$.
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
$endgroup$
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
add a comment |
$begingroup$
I think the cleanest way is$$sum_{substack Asubseteq [N]\|A|=n}Bigg(prod_{iin A}x_iBigg)Bigg(prod_{iin [n]setminus A}(1-x_i)Bigg),$$
where $[N]={1,2,dots,N}$.
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
$endgroup$
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
add a comment |
$begingroup$
I think the cleanest way is$$sum_{substack Asubseteq [N]\|A|=n}Bigg(prod_{iin A}x_iBigg)Bigg(prod_{iin [n]setminus A}(1-x_i)Bigg),$$
where $[N]={1,2,dots,N}$.
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
$endgroup$
I think the cleanest way is$$sum_{substack Asubseteq [N]\|A|=n}Bigg(prod_{iin A}x_iBigg)Bigg(prod_{iin [n]setminus A}(1-x_i)Bigg),$$
where $[N]={1,2,dots,N}$.
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
edited Jan 5 at 15:37
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
answered Jan 5 at 0:23
Mike EarnestMike Earnest
21.4k11951
21.4k11951
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
add a comment |
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
Thanks Mike, but I do not understand the relation between $[n]$ to the combinations. For example, for $N=4$ and $n=2$ the number of sum elements will be equal to ${4 choose 2}=3$. So did you mean that $[n]$ contains all combinations?
$endgroup$
– Gilsho
Jan 5 at 7:52
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
My bad, ${4 choose 2}=6$, but still more than $n=2$.
$endgroup$
– Gilsho
Jan 5 at 8:45
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
$begingroup$
I had some mistakes which are fixed now. $A$ ranges over all subsets of the numbers $1$ to $N$ of size $n$.
$endgroup$
– Mike Earnest
Jan 5 at 15:38
add a comment |
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$begingroup$
Let $X$ be your group of $N$ elements. Define $P$ (short for pairs) to be the set of ordered pairs $P = {(A,B)~:~Acap B = emptyset,~Acup B = X,~|A|=n}$. Your sum can then be written as $sumlimits_{(A,B)in P}left(prodlimits_{xin A}x~cdot~prodlimits_{xin B}(1-x)right)$. There are surely other ways to write this, but this is just one of the first to come to my mind.
$endgroup$
– JMoravitz
Jan 4 at 23:21