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












1














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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  • 1




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
















1














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


















  • 1




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














1












1








1


3





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













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











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asked Jan 4 at 15:47









user1942348user1942348

1,3731732




1,3731732








  • 1




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














  • 1




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








1




1




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




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










2 Answers
2






active

oldest

votes


















2














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





















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












  • @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.
    – Mike Earnest
    Jan 4 at 18:57










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



















1














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



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






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    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





















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












    • @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.
      – Mike Earnest
      Jan 4 at 18:57










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
















    2














    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





















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












    • @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.
      – Mike Earnest
      Jan 4 at 18:57










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














    2












    2








    2






    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












    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

    111k7106188




    111k7106188












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












    • @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.
      – Mike Earnest
      Jan 4 at 18:57










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


















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












    • @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.
      – Mike Earnest
      Jan 4 at 18:57










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
















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






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














    @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.
    – Mike Earnest
    Jan 4 at 18:57




    @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.
    – Mike Earnest
    Jan 4 at 18:57












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




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











    1














    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




























      1














      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


























        1












        1








        1






        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














        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

        3,148311




        3,148311






























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