Mixing
how many subsets in a set?
$begingroup$
I know this is a pretty basic question, but I fail to understand the notion behind it.
Consider the following:
If $A=left{a,b,cright}$, how many subsets can be created from $A$?
It is simple to write all possibilities and see it is $2^3=8$, but I don't understand why this is idea behind the formula.
combinatorics
asked Jan 7 at 19:06
segevpsegevp
586621
$endgroup$
add a comment |
$begingroup$
I know this is a pretty basic question, but I fail to understand the notion behind it.
Consider the following:
If $A=left{a,b,cright}$, how many subsets can be created from $A$?
It is simple to write all possibilities and see it is $2^3=8$, but I don't understand why this is idea behind the formula.
combinatorics
asked Jan 7 at 19:06
segevpsegevp
586621
$endgroup$
$begingroup$
you gave a set, not a group
$endgroup$
– mathworker21
Jan 7 at 19:10
$begingroup$
@mathworker21 my bad, I will edit.
$endgroup$
– segevp
Jan 7 at 19:11
1
$begingroup$
Think about it this way: Each element could be in the subgroup(really though, it makes more sense to say subset here), or it could not be in the subset. So for each element, there are two possibilities. Then we multiply all of the possibilities together, which is $(2)(2)(2) = 8.$ This works no matter how large the set is.
$endgroup$
– MathIsLife12
Jan 7 at 19:11
1
$begingroup$
@MathIsLife12 thanks, that is clear now
$endgroup$
– segevp
Jan 7 at 19:14
add a comment |
$begingroup$
I know this is a pretty basic question, but I fail to understand the notion behind it.
Consider the following:
If $A=left{a,b,cright}$, how many subsets can be created from $A$?
It is simple to write all possibilities and see it is $2^3=8$, but I don't understand why this is idea behind the formula.
combinatorics
asked Jan 7 at 19:06
segevpsegevp
586621
$endgroup$
I know this is a pretty basic question, but I fail to understand the notion behind it.
Consider the following:
If $A=left{a,b,cright}$, how many subsets can be created from $A$?
It is simple to write all possibilities and see it is $2^3=8$, but I don't understand why this is idea behind the formula.
combinatorics
combinatorics
asked Jan 7 at 19:06
segevpsegevp
586621
asked Jan 7 at 19:06
segevpsegevp
586621
edited Jan 7 at 19:12
segevp
asked Jan 7 at 19:06
segevpsegevp
586621
asked Jan 7 at 19:06
segevpsegevp
586621
asked Jan 7 at 19:06
segevpsegevp
586621
586621
$begingroup$
you gave a set, not a group
$endgroup$
– mathworker21
Jan 7 at 19:10
$begingroup$
@mathworker21 my bad, I will edit.
$endgroup$
– segevp
Jan 7 at 19:11
1
$begingroup$
Think about it this way: Each element could be in the subgroup(really though, it makes more sense to say subset here), or it could not be in the subset. So for each element, there are two possibilities. Then we multiply all of the possibilities together, which is $(2)(2)(2) = 8.$ This works no matter how large the set is.
$endgroup$
– MathIsLife12
Jan 7 at 19:11
1
$begingroup$
@MathIsLife12 thanks, that is clear now
$endgroup$
– segevp
Jan 7 at 19:14
add a comment |
$begingroup$
you gave a set, not a group
$endgroup$
– mathworker21
Jan 7 at 19:10
$begingroup$
@mathworker21 my bad, I will edit.
$endgroup$
– segevp
Jan 7 at 19:11
1
$begingroup$
Think about it this way: Each element could be in the subgroup(really though, it makes more sense to say subset here), or it could not be in the subset. So for each element, there are two possibilities. Then we multiply all of the possibilities together, which is $(2)(2)(2) = 8.$ This works no matter how large the set is.
$endgroup$
– MathIsLife12
Jan 7 at 19:11
1
$begingroup$
@MathIsLife12 thanks, that is clear now
$endgroup$
– segevp
Jan 7 at 19:14
$begingroup$
you gave a set, not a group
$endgroup$
– mathworker21
Jan 7 at 19:10
$begingroup$
you gave a set, not a group
$endgroup$
– mathworker21
Jan 7 at 19:10
$begingroup$
@mathworker21 my bad, I will edit.
$endgroup$
– segevp
Jan 7 at 19:11
$begingroup$
@mathworker21 my bad, I will edit.
$endgroup$
– segevp
Jan 7 at 19:11
1
1
$begingroup$
Think about it this way: Each element could be in the subgroup(really though, it makes more sense to say subset here), or it could not be in the subset. So for each element, there are two possibilities. Then we multiply all of the possibilities together, which is $(2)(2)(2) = 8.$ This works no matter how large the set is.
$endgroup$
– MathIsLife12
Jan 7 at 19:11
$begingroup$
Think about it this way: Each element could be in the subgroup(really though, it makes more sense to say subset here), or it could not be in the subset. So for each element, there are two possibilities. Then we multiply all of the possibilities together, which is $(2)(2)(2) = 8.$ This works no matter how large the set is.
$endgroup$
– MathIsLife12
Jan 7 at 19:11
1
1
$begingroup$
@MathIsLife12 thanks, that is clear now
$endgroup$
– segevp
Jan 7 at 19:14
$begingroup$
@MathIsLife12 thanks, that is clear now
$endgroup$
– segevp
Jan 7 at 19:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's say $A$ has $n$ elements (in your example $n=3$). To determine a subset of $A$, we go through each element of $A$ and decide whether to put it in our subset or to omit it from our subset. This creates two possible options for each element of $A$, which yields $2^n$ possible combinations once we go through all elements of $A$. Hence, $A$ has $2^n$ subsets.
answered Jan 7 at 19:14
DaveDave
8,78711033
$endgroup$
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let's say $A$ has $n$ elements (in your example $n=3$). To determine a subset of $A$, we go through each element of $A$ and decide whether to put it in our subset or to omit it from our subset. This creates two possible options for each element of $A$, which yields $2^n$ possible combinations once we go through all elements of $A$. Hence, $A$ has $2^n$ subsets.
answered Jan 7 at 19:14
DaveDave
8,78711033
$endgroup$
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
add a comment |
$begingroup$
Let's say $A$ has $n$ elements (in your example $n=3$). To determine a subset of $A$, we go through each element of $A$ and decide whether to put it in our subset or to omit it from our subset. This creates two possible options for each element of $A$, which yields $2^n$ possible combinations once we go through all elements of $A$. Hence, $A$ has $2^n$ subsets.
answered Jan 7 at 19:14
DaveDave
8,78711033
$endgroup$
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
add a comment |
$begingroup$
Let's say $A$ has $n$ elements (in your example $n=3$). To determine a subset of $A$, we go through each element of $A$ and decide whether to put it in our subset or to omit it from our subset. This creates two possible options for each element of $A$, which yields $2^n$ possible combinations once we go through all elements of $A$. Hence, $A$ has $2^n$ subsets.
answered Jan 7 at 19:14
DaveDave
8,78711033
$endgroup$
Let's say $A$ has $n$ elements (in your example $n=3$). To determine a subset of $A$, we go through each element of $A$ and decide whether to put it in our subset or to omit it from our subset. This creates two possible options for each element of $A$, which yields $2^n$ possible combinations once we go through all elements of $A$. Hence, $A$ has $2^n$ subsets.
answered Jan 7 at 19:14
DaveDave
8,78711033
answered Jan 7 at 19:14
DaveDave
8,78711033
answered Jan 7 at 19:14
DaveDave
8,78711033
answered Jan 7 at 19:14
DaveDave
8,78711033
8,78711033
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
add a comment |
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
Okay now I see it, this is the same as tossing a coin.
$endgroup$
– segevp
Jan 7 at 19:16
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
$begingroup$
In terms of the two outcomes for the elements, yes.
$endgroup$
– Dave
Jan 7 at 19:19
add a comment |
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$begingroup$
you gave a set, not a group
$endgroup$
– mathworker21
Jan 7 at 19:10
$begingroup$
@mathworker21 my bad, I will edit.
$endgroup$
– segevp
Jan 7 at 19:11
1
$begingroup$
Think about it this way: Each element could be in the subgroup(really though, it makes more sense to say subset here), or it could not be in the subset. So for each element, there are two possibilities. Then we multiply all of the possibilities together, which is $(2)(2)(2) = 8.$ This works no matter how large the set is.
$endgroup$
– MathIsLife12
Jan 7 at 19:11
1
$begingroup$
@MathIsLife12 thanks, that is clear now
$endgroup$
– segevp
Jan 7 at 19:14