Number of Subsets of $n$ Elements of Size $n-1$












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
    $endgroup$
    – Henry
    Jan 21 at 1:51
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    There are $2^n-n-1$ subsets which contain strictly fewer than $n−1$ elements
    $endgroup$
    – Henry
    Jan 21 at 1:51














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited Jan 21 at 1:31







Is12Prime

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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