How to prove $binom{n}{4}=sum_{i=3}^{n-1} binom{i}{3}$?












0












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I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics







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  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11
















0












$begingroup$


I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11














0












0








0





$begingroup$


I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics







share|cite|improve this question









$endgroup$




I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics




combinatorics






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asked Jan 15 at 14:00









Mahmud Nabil SnigdhoMahmud Nabil Snigdho

12




12












  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11


















  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11
















$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03






$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03






4




4




$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03




$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03












$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11




$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11










3 Answers
3






active

oldest

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3












$begingroup$

You may transform the sum into a telescoping sum using the "binomial" fact




  • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
      $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        You may transform the sum into a telescoping sum using the "binomial" fact




        • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


        Hence,
        begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
        & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
        & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
        & = & binom{n}{4}\
        end{eqnarray*}






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          You may transform the sum into a telescoping sum using the "binomial" fact




          • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


          Hence,
          begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
          & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
          & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
          & = & binom{n}{4}\
          end{eqnarray*}






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            You may transform the sum into a telescoping sum using the "binomial" fact




            • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


            Hence,
            begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
            & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
            & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
            & = & binom{n}{4}\
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$



            You may transform the sum into a telescoping sum using the "binomial" fact




            • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


            Hence,
            begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
            & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
            & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
            & = & binom{n}{4}\
            end{eqnarray*}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 15 at 14:15









            trancelocationtrancelocation

            11.9k1825




            11.9k1825























                3












                $begingroup$

                We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






                    share|cite|improve this answer









                    $endgroup$



                    We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 15 at 14:33









                    Christian BlatterChristian Blatter

                    174k8115327




                    174k8115327























                        3












                        $begingroup$

                        Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                        $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                          $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                            $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






                            share|cite|improve this answer









                            $endgroup$



                            Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                            $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 15 at 14:34









                            Marc van LeeuwenMarc van Leeuwen

                            87.4k5109224




                            87.4k5109224






























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