Mixing
Is it valid to define $binom{n}{n+k} = 0$
$begingroup$
Is it valid to define
$$binom{n}{n+k} = 0$$
where $k$ is an integer in ${k < -n}cup{k > n}$ ? I couldn't find anything on this notation via a quick google search, but I ran into it in the induction step of the proof that
$$sum_{k=0}^n{binom{n}{k}} = 2^n tag{$star$}$$
where the following is obtained by Pascal's identity:
begin{align}
sum_{k=0}^{n+1}{binom{n+1}{k}} &= sum_{k=0}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n+1}{binom{n}{k}} \ \
&= binom{n}{-1} + sum_{k=1}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n}{binom{n}{k}} + binom{n}{n+1} \ \
&= 0 + sum_{k=0}^{n}{binom{n}{k}} + sum_{k=0}^{n}{binom{n}{k}} + 0 \ \
&= 2cdot 2^n
end{align}
where this approach only makes sense if those aforementioned forms are zero. To me it seems to intuitively makes sense, since there are zero ways to do either of those things, since they are impossible.
I'm skeptical that this definition is valid because I saw a proof of $(star)$where those expressions were avoided by other reasoning.
combinatorics notation binomial-coefficients combinatorial-proofs
asked Feb 2 at 23:35
ZduffZduff
1,7501021
$endgroup$
add a comment |
$begingroup$
Is it valid to define
$$binom{n}{n+k} = 0$$
where $k$ is an integer in ${k < -n}cup{k > n}$ ? I couldn't find anything on this notation via a quick google search, but I ran into it in the induction step of the proof that
$$sum_{k=0}^n{binom{n}{k}} = 2^n tag{$star$}$$
where the following is obtained by Pascal's identity:
begin{align}
sum_{k=0}^{n+1}{binom{n+1}{k}} &= sum_{k=0}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n+1}{binom{n}{k}} \ \
&= binom{n}{-1} + sum_{k=1}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n}{binom{n}{k}} + binom{n}{n+1} \ \
&= 0 + sum_{k=0}^{n}{binom{n}{k}} + sum_{k=0}^{n}{binom{n}{k}} + 0 \ \
&= 2cdot 2^n
end{align}
where this approach only makes sense if those aforementioned forms are zero. To me it seems to intuitively makes sense, since there are zero ways to do either of those things, since they are impossible.
I'm skeptical that this definition is valid because I saw a proof of $(star)$where those expressions were avoided by other reasoning.
combinatorics notation binomial-coefficients combinatorial-proofs
asked Feb 2 at 23:35
ZduffZduff
1,7501021
$endgroup$
1
$begingroup$
See also en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
$endgroup$
– lhf
Feb 2 at 23:58
$begingroup$
How many ways are there to choose more than $n$ items from a collection of $n$ items?
$endgroup$
– amd
Feb 3 at 2:12
add a comment |
$begingroup$
Is it valid to define
$$binom{n}{n+k} = 0$$
where $k$ is an integer in ${k < -n}cup{k > n}$ ? I couldn't find anything on this notation via a quick google search, but I ran into it in the induction step of the proof that
$$sum_{k=0}^n{binom{n}{k}} = 2^n tag{$star$}$$
where the following is obtained by Pascal's identity:
begin{align}
sum_{k=0}^{n+1}{binom{n+1}{k}} &= sum_{k=0}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n+1}{binom{n}{k}} \ \
&= binom{n}{-1} + sum_{k=1}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n}{binom{n}{k}} + binom{n}{n+1} \ \
&= 0 + sum_{k=0}^{n}{binom{n}{k}} + sum_{k=0}^{n}{binom{n}{k}} + 0 \ \
&= 2cdot 2^n
end{align}
where this approach only makes sense if those aforementioned forms are zero. To me it seems to intuitively makes sense, since there are zero ways to do either of those things, since they are impossible.
I'm skeptical that this definition is valid because I saw a proof of $(star)$where those expressions were avoided by other reasoning.
combinatorics notation binomial-coefficients combinatorial-proofs
asked Feb 2 at 23:35
ZduffZduff
1,7501021
$endgroup$
Is it valid to define
$$binom{n}{n+k} = 0$$
where $k$ is an integer in ${k < -n}cup{k > n}$ ? I couldn't find anything on this notation via a quick google search, but I ran into it in the induction step of the proof that
$$sum_{k=0}^n{binom{n}{k}} = 2^n tag{$star$}$$
where the following is obtained by Pascal's identity:
begin{align}
sum_{k=0}^{n+1}{binom{n+1}{k}} &= sum_{k=0}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n+1}{binom{n}{k}} \ \
&= binom{n}{-1} + sum_{k=1}^{n+1}{binom{n}{k-1}} + sum_{k=0}^{n}{binom{n}{k}} + binom{n}{n+1} \ \
&= 0 + sum_{k=0}^{n}{binom{n}{k}} + sum_{k=0}^{n}{binom{n}{k}} + 0 \ \
&= 2cdot 2^n
end{align}
where this approach only makes sense if those aforementioned forms are zero. To me it seems to intuitively makes sense, since there are zero ways to do either of those things, since they are impossible.
I'm skeptical that this definition is valid because I saw a proof of $(star)$where those expressions were avoided by other reasoning.
combinatorics notation binomial-coefficients combinatorial-proofs
combinatorics notation binomial-coefficients combinatorial-proofs
asked Feb 2 at 23:35
ZduffZduff
1,7501021
asked Feb 2 at 23:35
ZduffZduff
1,7501021
asked Feb 2 at 23:35
ZduffZduff
1,7501021
asked Feb 2 at 23:35
ZduffZduff
1,7501021
asked Feb 2 at 23:35
ZduffZduff
1,7501021
1,7501021
1
$begingroup$
See also en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
$endgroup$
– lhf
Feb 2 at 23:58
$begingroup$
How many ways are there to choose more than $n$ items from a collection of $n$ items?
$endgroup$
– amd
Feb 3 at 2:12
add a comment |
1
$begingroup$
See also en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
$endgroup$
– lhf
Feb 2 at 23:58
$begingroup$
How many ways are there to choose more than $n$ items from a collection of $n$ items?
$endgroup$
– amd
Feb 3 at 2:12
1
1
$begingroup$
See also en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
$endgroup$
– lhf
Feb 2 at 23:58
$begingroup$
See also en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
$endgroup$
– lhf
Feb 2 at 23:58
$begingroup$
How many ways are there to choose more than $n$ items from a collection of $n$ items?
$endgroup$
– amd
Feb 3 at 2:12
$begingroup$
How many ways are there to choose more than $n$ items from a collection of $n$ items?
$endgroup$
– amd
Feb 3 at 2:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You may extend the definition of $n choose r$, by making it 0 if r<0 or r>n.
If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
$endgroup$
add a comment |
$begingroup$
It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.
The extended definition is
begin{align*}
binom{r}{k}=
begin{cases}
frac{r(r-1)cdots (r-k+1)}{k!}&qquad text{integer }kgeq 0\
0&qquad text{integer }k <0
end{cases}tag{1}
end{align*}
This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $binom{n}{n+k}=0$.
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You may extend the definition of $n choose r$, by making it 0 if r<0 or r>n.
If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
$endgroup$
add a comment |
$begingroup$
You may extend the definition of $n choose r$, by making it 0 if r<0 or r>n.
If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
$endgroup$
add a comment |
$begingroup$
You may extend the definition of $n choose r$, by making it 0 if r<0 or r>n.
If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
$endgroup$
You may extend the definition of $n choose r$, by making it 0 if r<0 or r>n.
If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
answered Feb 2 at 23:43
AlexandrosAlexandros
1,0701413
1,0701413
add a comment |
add a comment |
$begingroup$
It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.
The extended definition is
begin{align*}
binom{r}{k}=
begin{cases}
frac{r(r-1)cdots (r-k+1)}{k!}&qquad text{integer }kgeq 0\
0&qquad text{integer }k <0
end{cases}tag{1}
end{align*}
This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $binom{n}{n+k}=0$.
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
add a comment |
$begingroup$
It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.
The extended definition is
begin{align*}
binom{r}{k}=
begin{cases}
frac{r(r-1)cdots (r-k+1)}{k!}&qquad text{integer }kgeq 0\
0&qquad text{integer }k <0
end{cases}tag{1}
end{align*}
This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $binom{n}{n+k}=0$.
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
add a comment |
$begingroup$
It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.
The extended definition is
begin{align*}
binom{r}{k}=
begin{cases}
frac{r(r-1)cdots (r-k+1)}{k!}&qquad text{integer }kgeq 0\
0&qquad text{integer }k <0
end{cases}tag{1}
end{align*}
This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $binom{n}{n+k}=0$.
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
$endgroup$
It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.
The extended definition is
begin{align*}
binom{r}{k}=
begin{cases}
frac{r(r-1)cdots (r-k+1)}{k!}&qquad text{integer }kgeq 0\
0&qquad text{integer }k <0
end{cases}tag{1}
end{align*}
This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $binom{n}{n+k}=0$.
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
answered Feb 3 at 22:52
Markus ScheuerMarkus Scheuer
64.4k460152
64.4k460152
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
add a comment |
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
Fine, but why is this an "extension"? They are called "binomial coefficients", not "combination numbers". I always thought that the name "binomial coefficient" referred to their appearance as coefficients in the binomial expansion $(1+x)^m=sum_{n=0}^inftybinom mkx^k$ which is valid for any exponent $m$.
$endgroup$
– bof
Feb 6 at 9:16
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
$begingroup$
@bof: It's an extension to $binom{n}{k}=frac{n!}{k!(n-k)!}$ where $0leq kleq n$, $k,n$ integer.
$endgroup$
– Markus Scheuer
Feb 6 at 9:32
add a comment |
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1
$begingroup$
See also en.wikipedia.org/wiki/Pascal%27s_triangle#Extensions
$endgroup$
– lhf
Feb 2 at 23:58
$begingroup$
How many ways are there to choose more than $n$ items from a collection of $n$ items?
$endgroup$
– amd
Feb 3 at 2:12