simplify summation with binomial coefficients [closed]
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Is there a way to simplify this summation?:
$$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$
discrete-mathematics binomial-coefficients
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
asked Dec 22 '18 at 17:04
EkhEkh
92
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closed as off-topic by Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho Dec 23 '18 at 10:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
$begingroup$
Is there a way to simplify this summation?:
$$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$
discrete-mathematics binomial-coefficients
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
asked Dec 22 '18 at 17:04
EkhEkh
92
$endgroup$
closed as off-topic by Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho Dec 23 '18 at 10:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
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Have you tried expanding?
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– Sean Roberson
Dec 22 '18 at 17:16
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Have you tried oeis.org ?
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– Crostul
Dec 22 '18 at 17:48
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It can be simplified to $n!sum_{k=0}^n frac{1}{k!}binom{n}{k}$. I don't know if it can be evaluated.
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– herb steinberg
Dec 22 '18 at 17:50
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yes, I got that too in the form of $n!sum ^n _{k=0} frac{binom{n}{k}}{k!}$
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– Prakhar Nagpal
Dec 22 '18 at 17:51
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@Prakhar Nagpal Note correction.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:54
|
show 3 more comments
$begingroup$
Is there a way to simplify this summation?:
$$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$
discrete-mathematics binomial-coefficients
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
asked Dec 22 '18 at 17:04
EkhEkh
92
$endgroup$
Is there a way to simplify this summation?:
$$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$
discrete-mathematics binomial-coefficients
discrete-mathematics binomial-coefficients
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
asked Dec 22 '18 at 17:04
EkhEkh
92
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
asked Dec 22 '18 at 17:04
EkhEkh
92
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
edited Dec 22 '18 at 17:18
Prakhar Nagpal
747318
747318
asked Dec 22 '18 at 17:04
EkhEkh
92
asked Dec 22 '18 at 17:04
EkhEkh
92
asked Dec 22 '18 at 17:04
EkhEkh
92
92
closed as off-topic by Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho Dec 23 '18 at 10:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho Dec 23 '18 at 10:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, José Carlos Santos, amWhy, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Have you tried expanding?
$endgroup$
– Sean Roberson
Dec 22 '18 at 17:16
$begingroup$
Have you tried oeis.org ?
$endgroup$
– Crostul
Dec 22 '18 at 17:48
$begingroup$
It can be simplified to $n!sum_{k=0}^n frac{1}{k!}binom{n}{k}$. I don't know if it can be evaluated.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:50
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yes, I got that too in the form of $n!sum ^n _{k=0} frac{binom{n}{k}}{k!}$
$endgroup$
– Prakhar Nagpal
Dec 22 '18 at 17:51
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@Prakhar Nagpal Note correction.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:54
|
show 3 more comments
$begingroup$
Have you tried expanding?
$endgroup$
– Sean Roberson
Dec 22 '18 at 17:16
$begingroup$
Have you tried oeis.org ?
$endgroup$
– Crostul
Dec 22 '18 at 17:48
$begingroup$
It can be simplified to $n!sum_{k=0}^n frac{1}{k!}binom{n}{k}$. I don't know if it can be evaluated.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:50
$begingroup$
yes, I got that too in the form of $n!sum ^n _{k=0} frac{binom{n}{k}}{k!}$
$endgroup$
– Prakhar Nagpal
Dec 22 '18 at 17:51
$begingroup$
@Prakhar Nagpal Note correction.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:54
$begingroup$
Have you tried expanding?
$endgroup$
– Sean Roberson
Dec 22 '18 at 17:16
$begingroup$
Have you tried expanding?
$endgroup$
– Sean Roberson
Dec 22 '18 at 17:16
$begingroup$
Have you tried oeis.org ?
$endgroup$
– Crostul
Dec 22 '18 at 17:48
$begingroup$
Have you tried oeis.org ?
$endgroup$
– Crostul
Dec 22 '18 at 17:48
$begingroup$
It can be simplified to $n!sum_{k=0}^n frac{1}{k!}binom{n}{k}$. I don't know if it can be evaluated.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:50
$begingroup$
It can be simplified to $n!sum_{k=0}^n frac{1}{k!}binom{n}{k}$. I don't know if it can be evaluated.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:50
$begingroup$
yes, I got that too in the form of $n!sum ^n _{k=0} frac{binom{n}{k}}{k!}$
$endgroup$
– Prakhar Nagpal
Dec 22 '18 at 17:51
$begingroup$
yes, I got that too in the form of $n!sum ^n _{k=0} frac{binom{n}{k}}{k!}$
$endgroup$
– Prakhar Nagpal
Dec 22 '18 at 17:51
$begingroup$
@Prakhar Nagpal Note correction.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:54
$begingroup$
@Prakhar Nagpal Note correction.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:54
|
show 3 more comments
2 Answers
2
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oldest
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Alright, some of the comments have pointed out that it seems pretty difficult to find a definite formula but we can simplify it down to something much nicer to handle. We have, $$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$ Now, multiplying and dividing $k!$ and $n!$we get, $$n! sum^n _{k=0} frac{1}{k!} frac{left(n-kright)! cdot k!}{n!} cdotbinom{n}{k}^2 $$ This is equivalent to $$n! sum^n _{k=0} frac{binom{n}{k}^2}{binom{n}{k}k!}$$ Which is $$n! sum^n _{k=0} frac{binom n k}{k!}$$
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
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add a comment |
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The simplest form seems to be $n! L_n(-1)$, where $L_n(x)$ is a Laguerre polynomial.
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Alright, some of the comments have pointed out that it seems pretty difficult to find a definite formula but we can simplify it down to something much nicer to handle. We have, $$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$ Now, multiplying and dividing $k!$ and $n!$we get, $$n! sum^n _{k=0} frac{1}{k!} frac{left(n-kright)! cdot k!}{n!} cdotbinom{n}{k}^2 $$ This is equivalent to $$n! sum^n _{k=0} frac{binom{n}{k}^2}{binom{n}{k}k!}$$ Which is $$n! sum^n _{k=0} frac{binom n k}{k!}$$
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
$endgroup$
add a comment |
$begingroup$
Alright, some of the comments have pointed out that it seems pretty difficult to find a definite formula but we can simplify it down to something much nicer to handle. We have, $$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$ Now, multiplying and dividing $k!$ and $n!$we get, $$n! sum^n _{k=0} frac{1}{k!} frac{left(n-kright)! cdot k!}{n!} cdotbinom{n}{k}^2 $$ This is equivalent to $$n! sum^n _{k=0} frac{binom{n}{k}^2}{binom{n}{k}k!}$$ Which is $$n! sum^n _{k=0} frac{binom n k}{k!}$$
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
$endgroup$
add a comment |
$begingroup$
Alright, some of the comments have pointed out that it seems pretty difficult to find a definite formula but we can simplify it down to something much nicer to handle. We have, $$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$ Now, multiplying and dividing $k!$ and $n!$we get, $$n! sum^n _{k=0} frac{1}{k!} frac{left(n-kright)! cdot k!}{n!} cdotbinom{n}{k}^2 $$ This is equivalent to $$n! sum^n _{k=0} frac{binom{n}{k}^2}{binom{n}{k}k!}$$ Which is $$n! sum^n _{k=0} frac{binom n k}{k!}$$
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
$endgroup$
Alright, some of the comments have pointed out that it seems pretty difficult to find a definite formula but we can simplify it down to something much nicer to handle. We have, $$sum ^n _{k=0} left(n-kright)!binom{n}{k}^2$$ Now, multiplying and dividing $k!$ and $n!$we get, $$n! sum^n _{k=0} frac{1}{k!} frac{left(n-kright)! cdot k!}{n!} cdotbinom{n}{k}^2 $$ This is equivalent to $$n! sum^n _{k=0} frac{binom{n}{k}^2}{binom{n}{k}k!}$$ Which is $$n! sum^n _{k=0} frac{binom n k}{k!}$$
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
answered Dec 22 '18 at 18:02
Prakhar NagpalPrakhar Nagpal
747318
747318
add a comment |
add a comment |
$begingroup$
The simplest form seems to be $n! L_n(-1)$, where $L_n(x)$ is a Laguerre polynomial.
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
$endgroup$
add a comment |
$begingroup$
The simplest form seems to be $n! L_n(-1)$, where $L_n(x)$ is a Laguerre polynomial.
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
$endgroup$
add a comment |
$begingroup$
The simplest form seems to be $n! L_n(-1)$, where $L_n(x)$ is a Laguerre polynomial.
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
$endgroup$
The simplest form seems to be $n! L_n(-1)$, where $L_n(x)$ is a Laguerre polynomial.
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
answered Dec 22 '18 at 18:18
Pierpaolo VivoPierpaolo Vivo
5,3562724
5,3562724
add a comment |
add a comment |
$begingroup$
Have you tried expanding?
$endgroup$
– Sean Roberson
Dec 22 '18 at 17:16
$begingroup$
Have you tried oeis.org ?
$endgroup$
– Crostul
Dec 22 '18 at 17:48
$begingroup$
It can be simplified to $n!sum_{k=0}^n frac{1}{k!}binom{n}{k}$. I don't know if it can be evaluated.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:50
$begingroup$
yes, I got that too in the form of $n!sum ^n _{k=0} frac{binom{n}{k}}{k!}$
$endgroup$
– Prakhar Nagpal
Dec 22 '18 at 17:51
$begingroup$
@Prakhar Nagpal Note correction.
$endgroup$
– herb steinberg
Dec 22 '18 at 17:54