Number of Subsets of $n$ Elements of Size $n-1$
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I am trying to understand why given a set of $n$ elements, there are $2^{n-1}$ subsets of size $n-1$.
Below I have written out a proof that I figured out, however I was wondering if there was a simpler way of showing the result that might be a little more intuitive. Or, perhaps, a method that made use of combinatorics.
There are $2^n$ subsets of $n$ elements, $2^n -1 $ of which contain less than $n$ elements. Additionally, there are $2^{n-1} - 1$ subsets of n elements in which each set contains less than $n-1$ elements. Thus, the number of subsets of $n$ elements of size $n-1$ is: $(2^n -1) - (2^{n-1} -1) = 2^{n-1} $
probability combinatorics
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
$endgroup$
add a comment |
$begingroup$
I am trying to understand why given a set of $n$ elements, there are $2^{n-1}$ subsets of size $n-1$.
Below I have written out a proof that I figured out, however I was wondering if there was a simpler way of showing the result that might be a little more intuitive. Or, perhaps, a method that made use of combinatorics.
There are $2^n$ subsets of $n$ elements, $2^n -1 $ of which contain less than $n$ elements. Additionally, there are $2^{n-1} - 1$ subsets of n elements in which each set contains less than $n-1$ elements. Thus, the number of subsets of $n$ elements of size $n-1$ is: $(2^n -1) - (2^{n-1} -1) = 2^{n-1} $
probability combinatorics
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
$endgroup$
$begingroup$
There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
$endgroup$
– Henry
Jan 21 at 1:51
add a comment |
$begingroup$
I am trying to understand why given a set of $n$ elements, there are $2^{n-1}$ subsets of size $n-1$.
Below I have written out a proof that I figured out, however I was wondering if there was a simpler way of showing the result that might be a little more intuitive. Or, perhaps, a method that made use of combinatorics.
There are $2^n$ subsets of $n$ elements, $2^n -1 $ of which contain less than $n$ elements. Additionally, there are $2^{n-1} - 1$ subsets of n elements in which each set contains less than $n-1$ elements. Thus, the number of subsets of $n$ elements of size $n-1$ is: $(2^n -1) - (2^{n-1} -1) = 2^{n-1} $
probability combinatorics
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
$endgroup$
I am trying to understand why given a set of $n$ elements, there are $2^{n-1}$ subsets of size $n-1$.
Below I have written out a proof that I figured out, however I was wondering if there was a simpler way of showing the result that might be a little more intuitive. Or, perhaps, a method that made use of combinatorics.
There are $2^n$ subsets of $n$ elements, $2^n -1 $ of which contain less than $n$ elements. Additionally, there are $2^{n-1} - 1$ subsets of n elements in which each set contains less than $n-1$ elements. Thus, the number of subsets of $n$ elements of size $n-1$ is: $(2^n -1) - (2^{n-1} -1) = 2^{n-1} $
probability combinatorics
probability combinatorics
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
edited Jan 21 at 1:31
Is12Prime
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
asked Jan 21 at 0:38
Is12PrimeIs12Prime
130110
130110
$begingroup$
There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
$endgroup$
– Henry
Jan 21 at 1:51
add a comment |
$begingroup$
There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
$endgroup$
– Henry
Jan 21 at 1:51
$begingroup$
There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
$endgroup$
– Henry
Jan 21 at 1:51
$begingroup$
There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
$endgroup$
– Henry
Jan 21 at 1:51
add a comment |
1 Answer
1
active
oldest
votes
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It is not true. Given $n$ elements, you can select $n-1$ of them in ${n choose n-1}={n choose 1}=n$ ways, so there are $n$ subsets of size $n-1$. You can choose any element to leave out and you are done.
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
1
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
2
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
1
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
1
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
|
show 1 more comment
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1 Answer
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$begingroup$
It is not true. Given $n$ elements, you can select $n-1$ of them in ${n choose n-1}={n choose 1}=n$ ways, so there are $n$ subsets of size $n-1$. You can choose any element to leave out and you are done.
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
1
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
2
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
1
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
1
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
|
show 1 more comment
$begingroup$
It is not true. Given $n$ elements, you can select $n-1$ of them in ${n choose n-1}={n choose 1}=n$ ways, so there are $n$ subsets of size $n-1$. You can choose any element to leave out and you are done.
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
1
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
2
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
1
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
1
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
|
show 1 more comment
$begingroup$
It is not true. Given $n$ elements, you can select $n-1$ of them in ${n choose n-1}={n choose 1}=n$ ways, so there are $n$ subsets of size $n-1$. You can choose any element to leave out and you are done.
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
$endgroup$
It is not true. Given $n$ elements, you can select $n-1$ of them in ${n choose n-1}={n choose 1}=n$ ways, so there are $n$ subsets of size $n-1$. You can choose any element to leave out and you are done.
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
answered Jan 21 at 0:45
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
1
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
2
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
1
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
1
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
|
show 1 more comment
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
1
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
2
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
1
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
1
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
$begingroup$
This makes sense. However, I'm not sure what the error is in my false "proof" above.
$endgroup$
– Is12Prime
Jan 21 at 1:30
1
1
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
$begingroup$
@kelvinhong方 I would have thought there were $2^n-1$ (almost double $2^{n-1}$) subsets of size at most $n-1$
$endgroup$
– Henry
Jan 21 at 1:48
2
2
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
$begingroup$
@kelvinhong方: No, there are $2^n-n-1$ subsets of size at most $n-1$. There are $2^n$ total subsets, one is of size $n$, and $n$ are of size exactly $n-1$.
$endgroup$
– Ross Millikan
Jan 21 at 1:53
1
1
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
$begingroup$
@Is12Prime: The error in your proof is that for a given subset of $n-1$ elements there are $2^{n-1}-1$ subsets of size less than $n-1$, but there are $n$ subsets of size $n-1$. You then multiply count the smaller subsets because they can come from different subsets of size $n-1$
$endgroup$
– Ross Millikan
Jan 21 at 1:55
1
1
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
$begingroup$
@kelvinhong方: the last comment should have been there are $2^n-n-1$ subsets of size at most $n-2$. Yes, there are $2^n-1$ of size at most $n-1$ because only one is of size $n$.
$endgroup$
– Ross Millikan
Jan 21 at 2:03
|
show 1 more comment
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$begingroup$
There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
$endgroup$
– Henry
Jan 21 at 1:51