Explanation of counting by Inclusion Exclusion
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In my notes I have the following as an example for counting by inclusion exclusion.
Let S be a set. Let $c_i(x)$ where $i=1,2,3,4....k$, be a statement that is either true or false for $x in S$.
Denote $N(c_i)= |space {x in S| c_i(x)(is space true)}space|$
$N(c_ic_j)= |space {x in S| c_i(x) wedge}c_j(x)space |$
Then the number of elements of S that satisfy none of $c_1,c_2,c_3.....c_k$
is given by $|S|- sum_{1le ile k}N(c_i) + sum_{1le ile j le k}N(c_ic_j) - sum_{1le ile j le l le k}N(c_ic_jc_l) +........+(-1)^kN(c_1c_2c_3c_4......c_k)$
I am familiar enough with the Exclusion-Inclusion principle that I can express by understanding of it through a simple Venn diagram to convey how overcounting is being corrected. I honestly have no idea whatsoever what is happening in the above example. If someone could translate this into English I would greatly appreciate it. I've been at this for about an hour now so I am looking for a simple and really dumbed down explanation of what's going on.
combinatorics elementary-number-theory inclusion-exclusion
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In my notes I have the following as an example for counting by inclusion exclusion.
Let S be a set. Let $c_i(x)$ where $i=1,2,3,4....k$, be a statement that is either true or false for $x in S$.
Denote $N(c_i)= |space {x in S| c_i(x)(is space true)}space|$
$N(c_ic_j)= |space {x in S| c_i(x) wedge}c_j(x)space |$
Then the number of elements of S that satisfy none of $c_1,c_2,c_3.....c_k$
is given by $|S|- sum_{1le ile k}N(c_i) + sum_{1le ile j le k}N(c_ic_j) - sum_{1le ile j le l le k}N(c_ic_jc_l) +........+(-1)^kN(c_1c_2c_3c_4......c_k)$
I am familiar enough with the Exclusion-Inclusion principle that I can express by understanding of it through a simple Venn diagram to convey how overcounting is being corrected. I honestly have no idea whatsoever what is happening in the above example. If someone could translate this into English I would greatly appreciate it. I've been at this for about an hour now so I am looking for a simple and really dumbed down explanation of what's going on.
combinatorics elementary-number-theory inclusion-exclusion
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In my notes I have the following as an example for counting by inclusion exclusion.
Let S be a set. Let $c_i(x)$ where $i=1,2,3,4....k$, be a statement that is either true or false for $x in S$.
Denote $N(c_i)= |space {x in S| c_i(x)(is space true)}space|$
$N(c_ic_j)= |space {x in S| c_i(x) wedge}c_j(x)space |$
Then the number of elements of S that satisfy none of $c_1,c_2,c_3.....c_k$
is given by $|S|- sum_{1le ile k}N(c_i) + sum_{1le ile j le k}N(c_ic_j) - sum_{1le ile j le l le k}N(c_ic_jc_l) +........+(-1)^kN(c_1c_2c_3c_4......c_k)$
I am familiar enough with the Exclusion-Inclusion principle that I can express by understanding of it through a simple Venn diagram to convey how overcounting is being corrected. I honestly have no idea whatsoever what is happening in the above example. If someone could translate this into English I would greatly appreciate it. I've been at this for about an hour now so I am looking for a simple and really dumbed down explanation of what's going on.
combinatorics elementary-number-theory inclusion-exclusion
In my notes I have the following as an example for counting by inclusion exclusion.
Let S be a set. Let $c_i(x)$ where $i=1,2,3,4....k$, be a statement that is either true or false for $x in S$.
Denote $N(c_i)= |space {x in S| c_i(x)(is space true)}space|$
$N(c_ic_j)= |space {x in S| c_i(x) wedge}c_j(x)space |$
Then the number of elements of S that satisfy none of $c_1,c_2,c_3.....c_k$
is given by $|S|- sum_{1le ile k}N(c_i) + sum_{1le ile j le k}N(c_ic_j) - sum_{1le ile j le l le k}N(c_ic_jc_l) +........+(-1)^kN(c_1c_2c_3c_4......c_k)$
I am familiar enough with the Exclusion-Inclusion principle that I can express by understanding of it through a simple Venn diagram to convey how overcounting is being corrected. I honestly have no idea whatsoever what is happening in the above example. If someone could translate this into English I would greatly appreciate it. I've been at this for about an hour now so I am looking for a simple and really dumbed down explanation of what's going on.
combinatorics elementary-number-theory inclusion-exclusion
combinatorics elementary-number-theory inclusion-exclusion
asked Nov 24 at 5:40
user140161
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635614
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1 Answer
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The following is a somewhat more concrete statement of the principle, copied verbatim from Notes on Introductory Combinatorics by George Polya, Robert E. Tarjan and Donald R. Woods. I personally find this version very useful in actual applications.
"Suppose we have a set of $N$ objects that have various properties $alpha, beta, gamma, dots , lambda$. Each of the objects may have any or none of the properties. Let $N_alpha$ be the number of objects that have property $alpha$. Some of these objects may have other properties in addition to property $alpha$; that doesn't matter. (In fact, that's the whole idea!) Similarly, let $N_beta$ be the number of objects that have property $beta$, and so on. Let $N_{alpha beta}$ be the number of objects that have both property $alpha$ and property $beta$, $N_{alpha gamma}$ be the number that have properties $alpha$ and $gamma$, etc. $N_{alpha beta gamma dots lambda}$ is the number of objects with all the properties. Given this information, we want to find $N_0$, the number of objects that have none of the properties."
"The general formula for computing this is called the Principle of Inclusion and Exclusion (or sometimes PIE for short), and is the following:
$$begin{align}
N_0 = N &- N_alpha - N_beta - N_gamma - dots - N_lambda \
&+ N_{alpha beta} + N_{alpha gamma} + N_{beta gamma} + dots + N_{kappa lambda} \
&- N_{alpha beta gamma} - N_{alpha beta delta} - dots \
& vdots \
& pm N_{alpha beta gamma dots lambda}
end{align}$$
End of quotation.
In actual practice, I might add, the trickiest part may be formulating the problem to fit into this framework. Note that PIE requires that your final answer be the number of objects that have none of the properties. This forces you to "think negatively".
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The following is a somewhat more concrete statement of the principle, copied verbatim from Notes on Introductory Combinatorics by George Polya, Robert E. Tarjan and Donald R. Woods. I personally find this version very useful in actual applications.
"Suppose we have a set of $N$ objects that have various properties $alpha, beta, gamma, dots , lambda$. Each of the objects may have any or none of the properties. Let $N_alpha$ be the number of objects that have property $alpha$. Some of these objects may have other properties in addition to property $alpha$; that doesn't matter. (In fact, that's the whole idea!) Similarly, let $N_beta$ be the number of objects that have property $beta$, and so on. Let $N_{alpha beta}$ be the number of objects that have both property $alpha$ and property $beta$, $N_{alpha gamma}$ be the number that have properties $alpha$ and $gamma$, etc. $N_{alpha beta gamma dots lambda}$ is the number of objects with all the properties. Given this information, we want to find $N_0$, the number of objects that have none of the properties."
"The general formula for computing this is called the Principle of Inclusion and Exclusion (or sometimes PIE for short), and is the following:
$$begin{align}
N_0 = N &- N_alpha - N_beta - N_gamma - dots - N_lambda \
&+ N_{alpha beta} + N_{alpha gamma} + N_{beta gamma} + dots + N_{kappa lambda} \
&- N_{alpha beta gamma} - N_{alpha beta delta} - dots \
& vdots \
& pm N_{alpha beta gamma dots lambda}
end{align}$$
End of quotation.
In actual practice, I might add, the trickiest part may be formulating the problem to fit into this framework. Note that PIE requires that your final answer be the number of objects that have none of the properties. This forces you to "think negatively".
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
add a comment |
up vote
1
down vote
accepted
The following is a somewhat more concrete statement of the principle, copied verbatim from Notes on Introductory Combinatorics by George Polya, Robert E. Tarjan and Donald R. Woods. I personally find this version very useful in actual applications.
"Suppose we have a set of $N$ objects that have various properties $alpha, beta, gamma, dots , lambda$. Each of the objects may have any or none of the properties. Let $N_alpha$ be the number of objects that have property $alpha$. Some of these objects may have other properties in addition to property $alpha$; that doesn't matter. (In fact, that's the whole idea!) Similarly, let $N_beta$ be the number of objects that have property $beta$, and so on. Let $N_{alpha beta}$ be the number of objects that have both property $alpha$ and property $beta$, $N_{alpha gamma}$ be the number that have properties $alpha$ and $gamma$, etc. $N_{alpha beta gamma dots lambda}$ is the number of objects with all the properties. Given this information, we want to find $N_0$, the number of objects that have none of the properties."
"The general formula for computing this is called the Principle of Inclusion and Exclusion (or sometimes PIE for short), and is the following:
$$begin{align}
N_0 = N &- N_alpha - N_beta - N_gamma - dots - N_lambda \
&+ N_{alpha beta} + N_{alpha gamma} + N_{beta gamma} + dots + N_{kappa lambda} \
&- N_{alpha beta gamma} - N_{alpha beta delta} - dots \
& vdots \
& pm N_{alpha beta gamma dots lambda}
end{align}$$
End of quotation.
In actual practice, I might add, the trickiest part may be formulating the problem to fit into this framework. Note that PIE requires that your final answer be the number of objects that have none of the properties. This forces you to "think negatively".
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The following is a somewhat more concrete statement of the principle, copied verbatim from Notes on Introductory Combinatorics by George Polya, Robert E. Tarjan and Donald R. Woods. I personally find this version very useful in actual applications.
"Suppose we have a set of $N$ objects that have various properties $alpha, beta, gamma, dots , lambda$. Each of the objects may have any or none of the properties. Let $N_alpha$ be the number of objects that have property $alpha$. Some of these objects may have other properties in addition to property $alpha$; that doesn't matter. (In fact, that's the whole idea!) Similarly, let $N_beta$ be the number of objects that have property $beta$, and so on. Let $N_{alpha beta}$ be the number of objects that have both property $alpha$ and property $beta$, $N_{alpha gamma}$ be the number that have properties $alpha$ and $gamma$, etc. $N_{alpha beta gamma dots lambda}$ is the number of objects with all the properties. Given this information, we want to find $N_0$, the number of objects that have none of the properties."
"The general formula for computing this is called the Principle of Inclusion and Exclusion (or sometimes PIE for short), and is the following:
$$begin{align}
N_0 = N &- N_alpha - N_beta - N_gamma - dots - N_lambda \
&+ N_{alpha beta} + N_{alpha gamma} + N_{beta gamma} + dots + N_{kappa lambda} \
&- N_{alpha beta gamma} - N_{alpha beta delta} - dots \
& vdots \
& pm N_{alpha beta gamma dots lambda}
end{align}$$
End of quotation.
In actual practice, I might add, the trickiest part may be formulating the problem to fit into this framework. Note that PIE requires that your final answer be the number of objects that have none of the properties. This forces you to "think negatively".
The following is a somewhat more concrete statement of the principle, copied verbatim from Notes on Introductory Combinatorics by George Polya, Robert E. Tarjan and Donald R. Woods. I personally find this version very useful in actual applications.
"Suppose we have a set of $N$ objects that have various properties $alpha, beta, gamma, dots , lambda$. Each of the objects may have any or none of the properties. Let $N_alpha$ be the number of objects that have property $alpha$. Some of these objects may have other properties in addition to property $alpha$; that doesn't matter. (In fact, that's the whole idea!) Similarly, let $N_beta$ be the number of objects that have property $beta$, and so on. Let $N_{alpha beta}$ be the number of objects that have both property $alpha$ and property $beta$, $N_{alpha gamma}$ be the number that have properties $alpha$ and $gamma$, etc. $N_{alpha beta gamma dots lambda}$ is the number of objects with all the properties. Given this information, we want to find $N_0$, the number of objects that have none of the properties."
"The general formula for computing this is called the Principle of Inclusion and Exclusion (or sometimes PIE for short), and is the following:
$$begin{align}
N_0 = N &- N_alpha - N_beta - N_gamma - dots - N_lambda \
&+ N_{alpha beta} + N_{alpha gamma} + N_{beta gamma} + dots + N_{kappa lambda} \
&- N_{alpha beta gamma} - N_{alpha beta delta} - dots \
& vdots \
& pm N_{alpha beta gamma dots lambda}
end{align}$$
End of quotation.
In actual practice, I might add, the trickiest part may be formulating the problem to fit into this framework. Note that PIE requires that your final answer be the number of objects that have none of the properties. This forces you to "think negatively".
edited Nov 24 at 16:09
answered Nov 24 at 16:01
awkward
5,71111022
5,71111022
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
add a comment |
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
What does N denote here? The number of all possible objects? Could you give a concrete example of this?
– user140161
Nov 24 at 19:03
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
@user140161 Yes, $N$ is the number of all objects. One of my solutions using PIE is math.stackexchange.com/questions/2032110/…. Another is math.stackexchange.com/questions/2879280/…
– awkward
Nov 24 at 20:23
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think I get it now. So for example if you have sets ABC in a universe U, then what we are really doing is counting ABC by PIE, and then subtracting that from U to calculate $(ABC)^c$. It's kind of like a complement version of PIE
– user140161
Nov 24 at 20:46
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
I think the only confusion I have left is why the last term has $(-1)^k$
– user140161
Nov 24 at 20:47
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
@user140161 Suppose you have two sets $A$ and $B$; then $|A cup B| = |A| + |B| - |A cap B|$. If you add the sizes of two sets together, you've counted the stuff in their intersection twice, so you have to subtract it out. PIE is just the generalization to more than two sets.
– awkward
Nov 25 at 12:50
add a comment |
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