Prove that $sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} {^iC_j}=3^n$












1












$begingroup$


Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations







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$endgroup$








  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53
















1












$begingroup$


Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53














1












1








1


3



$begingroup$


Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations







share|cite|improve this question









$endgroup$




Prove that



$$sum_{j=0}^{n}sum_{i=j}^{n} {^nC_i} ,{^iC_j}=3^n$$



I have tried by interchanging j by i but unable to prove. Don't know which combination formula would be used. Please help.






combinatorics discrete-mathematics combinations




combinatorics discrete-mathematics combinations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 15:47









user1942348user1942348

1,3901934




1,3901934








  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53














  • 1




    $begingroup$
    Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
    $endgroup$
    – darij grinberg
    Jan 4 at 15:53








1




1




$begingroup$
Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
$endgroup$
– darij grinberg
Jan 4 at 15:53




$begingroup$
Hint: $dbinom{n}{i}dbinom{i}{j} = dbinom{n}{j} dbinom{n-j}{i-j}$.
$endgroup$
– darij grinberg
Jan 4 at 15:53










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hints:




  • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

  • $3=1+2$

  • $2=1+1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
    $endgroup$
    – user1942348
    Jan 4 at 16:53












  • $begingroup$
    @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
    $endgroup$
    – Mike Earnest
    Jan 4 at 18:57










  • $begingroup$
    @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
    $endgroup$
    – Arthur
    Jan 4 at 18:59



















1












$begingroup$

First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




  1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

  2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


(So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



$$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59
















    2












    $begingroup$

    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59














    2












    2








    2





    $begingroup$

    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$






    share|cite|improve this answer









    $endgroup$



    Hints:




    • $displaystyle sum_{j=0}^nsum_{i=j}^n=sum_{i=0}^nsum_{j=0}^i$

    • $3=1+2$

    • $2=1+1$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 15:58









    ArthurArthur

    122k7122211




    122k7122211












    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59


















    • $begingroup$
      Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
      $endgroup$
      – user1942348
      Jan 4 at 16:53












    • $begingroup$
      @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
      $endgroup$
      – Mike Earnest
      Jan 4 at 18:57










    • $begingroup$
      @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
      $endgroup$
      – Arthur
      Jan 4 at 18:59
















    $begingroup$
    Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
    $endgroup$
    – user1942348
    Jan 4 at 16:53






    $begingroup$
    Would you please write a bit more. How 3=1+2 and 2=1+1 helps?
    $endgroup$
    – user1942348
    Jan 4 at 16:53














    $begingroup$
    @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
    $endgroup$
    – Mike Earnest
    Jan 4 at 18:57




    $begingroup$
    @user1942348 After switching the order of summation, you can factor ${}^nC_i$ out of the inner summation. The remaining inner summation can be simplified with the binomial theorem.
    $endgroup$
    – Mike Earnest
    Jan 4 at 18:57












    $begingroup$
    @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
    $endgroup$
    – Arthur
    Jan 4 at 18:59




    $begingroup$
    @user1942348 $3^n=(1+2)^n$. Now use the binomial theorem.
    $endgroup$
    – Arthur
    Jan 4 at 18:59











    1












    $begingroup$

    First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




    1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

    2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


    (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



    On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



    $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




      1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

      2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


      (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



      On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



      $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




        1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

        2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


        (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



        On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



        $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$






        share|cite|improve this answer











        $endgroup$



        First, let $S_{ij}$ be the set of words $s$ in ${0,1,2}^n$ that satisfy both of the following 1. and 2. below:




        1. Precisely $n-i$ of the letters of $s$ are "0", oe equivalently, precisely $i$ letters of $s$ are in ${1,2}$.

        2. Of the $i$ letters that are in ${1,2}$, there are precsiely $j$ of those letters that are "2".


        (So $n-i$ letters of each $s in S_{ij}$ are "0", $i-j$ letters of $s$ are "1", and the remaining $j$ letters of $s$ are "2".)



        On the one hand: $|S_{ij}|= {^nC_i} ,{^iC_j}$; first choose the $(n-i)$ out of $n$ positions of $s$ such that the letter in each such position is "0", and then from the remaining $i$ positions, choose the $j$ positions such that the letter in each such position is a "2". On the other hand, the $S_{ij}$s partition ${0,1,2}^n$, which has cardinality $3^n$. So this gives



        $$3^n = sum_{i=1}^n sum_{j=1}^i |S_{ij}| = sum_{j=1}^n sum_{i=j}^n |S_{ij}| = sum_{j=1}^nsum_{i=j}^n {^nC_i} ,{^iC_j}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 19:52

























        answered Jan 4 at 19:31









        MikeMike

        4,611512




        4,611512






























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