Proving $binom{n}{k}=binom{n}{n-k}$ using set theory?












1












$begingroup$


Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.



Let $A$ be the set of all groupings of $k$ items from $N$.



Let $B$ be the set of all groupings of $n-k$ items from $N$.



Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.



For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$



We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.



We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.



Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.



Is this proof correct?










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  • $begingroup$
    It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
    $endgroup$
    – bof
    Jan 11 at 4:39
















1












$begingroup$


Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.



Let $A$ be the set of all groupings of $k$ items from $N$.



Let $B$ be the set of all groupings of $n-k$ items from $N$.



Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.



For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$



We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.



We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.



Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.



Is this proof correct?










share|cite|improve this question









$endgroup$












  • $begingroup$
    It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
    $endgroup$
    – bof
    Jan 11 at 4:39














1












1








1





$begingroup$


Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.



Let $A$ be the set of all groupings of $k$ items from $N$.



Let $B$ be the set of all groupings of $n-k$ items from $N$.



Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.



For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$



We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.



We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.



Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.



Is this proof correct?










share|cite|improve this question









$endgroup$




Let $N$ be a set of $n$ items. Define a grouping to be some subset of $N$.



Let $A$ be the set of all groupings of $k$ items from $N$.



Let $B$ be the set of all groupings of $n-k$ items from $N$.



Namely, if $|A|=|B|$ then we have $binom{n}{k}=binom{n}{n-k}$.



For $|A|=|B|$, we must have there exists a bijection from $A$ to $B$. Let us define a function $f:A rightarrow B$ such that $f(a)=b$ if $a cup b = N$



We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $a_1 cup b = N = a_2 cup b$, i.e. $a_1=a_2=N-b$. So, $f$ is injective.



We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $acup b=(N-b)cup b=b$.



Since $f$ is injective and surjective, it is bijective. So $|A|=|B|$, and as such we have $binom{n}{k}=binom{n}{n-k}$ QED.



Is this proof correct?







combinatorics proof-verification elementary-set-theory






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asked Jan 11 at 0:16









Hisoka MorohHisoka Moroh

420110




420110












  • $begingroup$
    It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
    $endgroup$
    – bof
    Jan 11 at 4:39


















  • $begingroup$
    It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
    $endgroup$
    – bof
    Jan 11 at 4:39
















$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39




$begingroup$
It would be hard to tell if your proof is correct without knowing what definition of $binom nk$ you are working with, and what properties of $binom nk$ have already been established. By the way, why do you define the word "grouping" to mean "set"? What's wrong with using the word "set" to mean "set"? Usually we introduce definitions to make things shorter, not to make them longer.
$endgroup$
– bof
Jan 11 at 4:39










1 Answer
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$begingroup$

The problem in this proof is with the definition of $f(a)$:




$f(a)=b$ if $a cup b = N$




However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:




$f(a)=b$ if $b=N-a$




Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.



Using this new definition, I will now modify your proof as follows:




We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.



We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.







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    $begingroup$

    The problem in this proof is with the definition of $f(a)$:




    $f(a)=b$ if $a cup b = N$




    However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:




    $f(a)=b$ if $b=N-a$




    Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.



    Using this new definition, I will now modify your proof as follows:




    We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.



    We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.







    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The problem in this proof is with the definition of $f(a)$:




      $f(a)=b$ if $a cup b = N$




      However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:




      $f(a)=b$ if $b=N-a$




      Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.



      Using this new definition, I will now modify your proof as follows:




      We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.



      We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.







      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The problem in this proof is with the definition of $f(a)$:




        $f(a)=b$ if $a cup b = N$




        However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:




        $f(a)=b$ if $b=N-a$




        Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.



        Using this new definition, I will now modify your proof as follows:




        We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.



        We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.







        share|cite|improve this answer









        $endgroup$



        The problem in this proof is with the definition of $f(a)$:




        $f(a)=b$ if $a cup b = N$




        However, how do we even know that such a $bin B$ exists? The fact that this function is well-defined must be proven rigorously. Therefore, I think this alternative definition would be better:




        $f(a)=b$ if $b=N-a$




        Now, we will prove $b=N-a in B$: Since $ain A$, we know that $|a|=k$. Furthermore, it is given that $|N|=n$. Therefore, since $a$ and $N$ are finite sets, $|N-a|=|N|-|a|=n-k$. Thus, $|b|=|N-a|=n-k$, so $b in B$. This proves that $f(a)=N-a$ is a well-defined function from $A$ to $B$.



        Using this new definition, I will now modify your proof as follows:




        We first show $f$ is injective. Take $a_1,a_2 in A$ and assume $f(a_1)=f(a_2)=b$. This means $b=N-a_1$ and $b=N-a_2$. Thus, $a_1=N-b$ and $a_2=N-b$. Therefore, $a_1=a_2=N-b$. So, $f$ is injective.



        We now show $f$ is surjective. Take $b in B$ and consider $a=N-b$. Since $|N-b|=n-(n-k)=k$, we have $|a|=k$ so $a in A$. We note $f(a)=b$ because $a=N-b$ implies $b=N-a$.








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        share|cite|improve this answer










        answered Jan 11 at 0:22









        Noble MushtakNoble Mushtak

        15.3k1835




        15.3k1835






























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