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Sum of product of certain binomial coefficients: $sumlimits_m = 0^k sumlimits_n = 0^m binomkm binom mn$

Prove the identity $sum_k=0^nsum_r=0^k binomkr binomnk = 3^n$A Binomial Coefficient Sum: $sum_m = 0^n (-1)^n-m binomnm binomm-1l$Upper bound on sum of binomial coefficientsSum involving integer compositions and binomial coefficientsCompute the value of a sum (as an expression involving one or two binomial coefficients)A sum with binomial coefficients in the numerator and denominator.Summation involving binomial coefficients: $sum_i=0^100 binom3003i$Sum of product of squared binomial coefficientsAlternating sum of binomial coefficients up to certain valueEvaluate the Binomial SumIs there a closed form for the sum of the cubes of the binomial coefficients?

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3

$begingroup$

Given a nonnegative integer $k$. What is the value of the following sum:
$$sum_m = 0^k sum_n = 0^m binomkm binommn$$

I need this in order to simplify some of my work. I tried to expand it using binomial formula but didn't lead anywhere. I am simplifying weyl denominator formula for certain Kac-Moody algebras and I end up with this sum.

summation binomial-coefficients

share|cite|improve this question

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Thomas Andrews
  Jul 14 at 1:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I expand it using binomial formula but didn't lead anywhere. I am simplifying weyl denominator formula for certain Kac-Moody algebras and I end up with this sum. So definitely it is not a homework sum. so please help me with this.
  $endgroup$
  – GA316
  Jul 14 at 1:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Prove the identity $sum_k=0^nsum_r=0^k binomkr binomnk = 3^n$
  $endgroup$
  – YuiTo Cheng
  Jul 20 at 4:51
  

  
  
  
  

add a comment | 

3

$begingroup$

Given a nonnegative integer $k$. What is the value of the following sum:
$$sum_m = 0^k sum_n = 0^m binomkm binommn$$

I need this in order to simplify some of my work. I tried to expand it using binomial formula but didn't lead anywhere. I am simplifying weyl denominator formula for certain Kac-Moody algebras and I end up with this sum.

summation binomial-coefficients

share|cite|improve this question

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Thomas Andrews
  Jul 14 at 1:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I expand it using binomial formula but didn't lead anywhere. I am simplifying weyl denominator formula for certain Kac-Moody algebras and I end up with this sum. So definitely it is not a homework sum. so please help me with this.
  $endgroup$
  – GA316
  Jul 14 at 1:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Prove the identity $sum_k=0^nsum_r=0^k binomkr binomnk = 3^n$
  $endgroup$
  – YuiTo Cheng
  Jul 20 at 4:51
  

  
  
  
  

add a comment | 

$begingroup$

Given a nonnegative integer $k$. What is the value of the following sum:
$$sum_m = 0^k sum_n = 0^m binomkm binommn$$

I need this in order to simplify some of my work. I tried to expand it using binomial formula but didn't lead anywhere. I am simplifying weyl denominator formula for certain Kac-Moody algebras and I end up with this sum.

summation binomial-coefficients

share|cite|improve this question

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

share|cite|improve this question

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

edited Jul 20 at 18:19

Martin Sleziak

46k1111 gold badges127127 silver badges283283 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

asked Jul 14 at 1:23

GA316GA316

2,7791414 silver badges3333 bronze badges

3 Answers
3

active

oldest

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4

$begingroup$

$$kchoose mm choose n = k!over m!(k-m)!m!over n!(m-n)!=kchoose k-m,n,m-n$$ so we have $$sum_m=0^ksum_n=0^mkchoose k-m,n,m-n=3^k,$$ the number of ways to distribute $k$ objects in $3$ piles.

share|cite|improve this answer

answered Jul 14 at 1:47

saulspatzsaulspatz

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add a comment | 

1

$begingroup$

The sum $$S=sum_m=0^k sum_n=0^m k choose mm choose n=sum_m=0^k k choose msum_n=0^m m choose n= sum_m=0^k 2^m k choose m=3^k.$$

share|cite|improve this answer

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

The answer given by @saulspatz is correct. I would like to give an alternative way to get to the result that helps understanding what is actually going on.

---

The binomial coefficient $binomkm$ is the number of ways in which we can choose $m$ elements out of a set of $k$ elements. In other words, it is the number of elements of the set
$$f:x_1,ldots,x_kto1,2mid #f^-1(2)=m.$$
Of course, we have a similar interpretation for $binommn$. Thus,
beginalign
binomkmbinommn =& #f:x_1,ldots,x_kto0,1mid #f^-1(0)=mcdot#g:f^-1(0)to2,3mid #g^-1(3)=n\
=&#(f,g)mid#f^-1(0)=m, #g^-1(3)=n
endalign
(I've omitted the domains and codomains of the maps in the last line). Now, given two such maps we can construct
$$h:x_1,ldots,x_klongrightarrow1,2,3$$
as the map given by $f$ on $f^-1(1)$ and $gcirc f$ otherwise, thus giving
$$binomkmbinommn=#h:x_1,ldots,x_kto1,2,3mid h^-1(3)=n, h^-1(2)=m-n .$$
Taking the sum over $m$ and $n$ is given by taking the union on the sets we are counting, which eliminates the conditions giving
$$sum_m=0^ksum_n=0^mbinomkmbinommn = #h:x_1,ldots,x_kto1,2,3 = 3^k.$$

share|cite|improve this answer

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

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$endgroup$

add a comment | 

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4

$begingroup$

$$kchoose mm choose n = k!over m!(k-m)!m!over n!(m-n)!=kchoose k-m,n,m-n$$ so we have $$sum_m=0^ksum_n=0^mkchoose k-m,n,m-n=3^k,$$ the number of ways to distribute $k$ objects in $3$ piles.

share|cite|improve this answer

answered Jul 14 at 1:47

saulspatzsaulspatz

21.9k44 gold badges1616 silver badges3838 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

$$kchoose mm choose n = k!over m!(k-m)!m!over n!(m-n)!=kchoose k-m,n,m-n$$ so we have $$sum_m=0^ksum_n=0^mkchoose k-m,n,m-n=3^k,$$ the number of ways to distribute $k$ objects in $3$ piles.

share|cite|improve this answer

answered Jul 14 at 1:47

saulspatzsaulspatz

21.9k44 gold badges1616 silver badges3838 bronze badges

$endgroup$

add a comment | 

$begingroup$

$$kchoose mm choose n = k!over m!(k-m)!m!over n!(m-n)!=kchoose k-m,n,m-n$$ so we have $$sum_m=0^ksum_n=0^mkchoose k-m,n,m-n=3^k,$$ the number of ways to distribute $k$ objects in $3$ piles.

share|cite|improve this answer

answered Jul 14 at 1:47

saulspatzsaulspatz

21.9k44 gold badges1616 silver badges3838 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 14 at 1:47

saulspatzsaulspatz

21.9k44 gold badges1616 silver badges3838 bronze badges

answered Jul 14 at 1:47

saulspatzsaulspatz

21.9k44 gold badges1616 silver badges3838 bronze badges

answered Jul 14 at 1:47

saulspatzsaulspatz

21.9k44 gold badges1616 silver badges3838 bronze badges

1

$begingroup$

The sum $$S=sum_m=0^k sum_n=0^m k choose mm choose n=sum_m=0^k k choose msum_n=0^m m choose n= sum_m=0^k 2^m k choose m=3^k.$$

share|cite|improve this answer

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

The sum $$S=sum_m=0^k sum_n=0^m k choose mm choose n=sum_m=0^k k choose msum_n=0^m m choose n= sum_m=0^k 2^m k choose m=3^k.$$

share|cite|improve this answer

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

$endgroup$

add a comment | 

$begingroup$

The sum $$S=sum_m=0^k sum_n=0^m k choose mm choose n=sum_m=0^k k choose msum_n=0^m m choose n= sum_m=0^k 2^m k choose m=3^k.$$

share|cite|improve this answer

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

answered Jul 14 at 2:26

Dr Zafar Ahmed DScDr Zafar Ahmed DSc

3,89533 silver badges1616 bronze badges

0

$begingroup$

The answer given by @saulspatz is correct. I would like to give an alternative way to get to the result that helps understanding what is actually going on.

---

The binomial coefficient $binomkm$ is the number of ways in which we can choose $m$ elements out of a set of $k$ elements. In other words, it is the number of elements of the set
$$f:x_1,ldots,x_kto1,2mid #f^-1(2)=m.$$
Of course, we have a similar interpretation for $binommn$. Thus,
beginalign
binomkmbinommn =& #f:x_1,ldots,x_kto0,1mid #f^-1(0)=mcdot#g:f^-1(0)to2,3mid #g^-1(3)=n\
=&#(f,g)mid#f^-1(0)=m, #g^-1(3)=n
endalign
(I've omitted the domains and codomains of the maps in the last line). Now, given two such maps we can construct
$$h:x_1,ldots,x_klongrightarrow1,2,3$$
as the map given by $f$ on $f^-1(1)$ and $gcirc f$ otherwise, thus giving
$$binomkmbinommn=#h:x_1,ldots,x_kto1,2,3mid h^-1(3)=n, h^-1(2)=m-n .$$
Taking the sum over $m$ and $n$ is given by taking the union on the sets we are counting, which eliminates the conditions giving
$$sum_m=0^ksum_n=0^mbinomkmbinommn = #h:x_1,ldots,x_kto1,2,3 = 3^k.$$

share|cite|improve this answer

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

20.9k33 gold badges3939 silver badges9898 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

The answer given by @saulspatz is correct. I would like to give an alternative way to get to the result that helps understanding what is actually going on.

---

The binomial coefficient $binomkm$ is the number of ways in which we can choose $m$ elements out of a set of $k$ elements. In other words, it is the number of elements of the set
$$f:x_1,ldots,x_kto1,2mid #f^-1(2)=m.$$
Of course, we have a similar interpretation for $binommn$. Thus,
beginalign
binomkmbinommn =& #f:x_1,ldots,x_kto0,1mid #f^-1(0)=mcdot#g:f^-1(0)to2,3mid #g^-1(3)=n\
=&#(f,g)mid#f^-1(0)=m, #g^-1(3)=n
endalign
(I've omitted the domains and codomains of the maps in the last line). Now, given two such maps we can construct
$$h:x_1,ldots,x_klongrightarrow1,2,3$$
as the map given by $f$ on $f^-1(1)$ and $gcirc f$ otherwise, thus giving
$$binomkmbinommn=#h:x_1,ldots,x_kto1,2,3mid h^-1(3)=n, h^-1(2)=m-n .$$
Taking the sum over $m$ and $n$ is given by taking the union on the sets we are counting, which eliminates the conditions giving
$$sum_m=0^ksum_n=0^mbinomkmbinommn = #h:x_1,ldots,x_kto1,2,3 = 3^k.$$

share|cite|improve this answer

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

20.9k33 gold badges3939 silver badges9898 bronze badges

$endgroup$

add a comment | 

$begingroup$

The answer given by @saulspatz is correct. I would like to give an alternative way to get to the result that helps understanding what is actually going on.

---

The binomial coefficient $binomkm$ is the number of ways in which we can choose $m$ elements out of a set of $k$ elements. In other words, it is the number of elements of the set
$$f:x_1,ldots,x_kto1,2mid #f^-1(2)=m.$$
Of course, we have a similar interpretation for $binommn$. Thus,
beginalign
binomkmbinommn =& #f:x_1,ldots,x_kto0,1mid #f^-1(0)=mcdot#g:f^-1(0)to2,3mid #g^-1(3)=n\
=&#(f,g)mid#f^-1(0)=m, #g^-1(3)=n
endalign
(I've omitted the domains and codomains of the maps in the last line). Now, given two such maps we can construct
$$h:x_1,ldots,x_klongrightarrow1,2,3$$
as the map given by $f$ on $f^-1(1)$ and $gcirc f$ otherwise, thus giving
$$binomkmbinommn=#h:x_1,ldots,x_kto1,2,3mid h^-1(3)=n, h^-1(2)=m-n .$$
Taking the sum over $m$ and $n$ is given by taking the union on the sets we are counting, which eliminates the conditions giving
$$sum_m=0^ksum_n=0^mbinomkmbinommn = #h:x_1,ldots,x_kto1,2,3 = 3^k.$$

share|cite|improve this answer

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

20.9k33 gold badges3939 silver badges9898 bronze badges

$endgroup$

share|cite|improve this answer

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

20.9k33 gold badges3939 silver badges9898 bronze badges

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

edited Jul 20 at 20:46

darij grinberg

12.2k44 gold badges3232 silver badges7272 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

20.9k33 gold badges3939 silver badges9898 bronze badges

answered Jul 14 at 10:40

Daniel Robert-NicoudDaniel Robert-Nicoud

20.9k33 gold badges3939 silver badges9898 bronze badges