Sum of Combination. Calculate $sum_k=0^10 binom10k^2$ [closed]

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I wanna know the answer of this problem and solve it:



$$binom100^2+binom101^2+binom102^2+cdots+binom1010^2 = ?$$



Can you help me? Thanks in advance.




power-series combinations




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closed as off-topic by asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel Jul 23 at 12:30


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Are you asking for a general solution for $sum_k=0^n binomnk^2$ or only for the special case where $n=10$?
    – mrtaurho
    Jul 23 at 8:08















up vote
-2
down vote

favorite












I wanna know the answer of this problem and solve it:



$$binom100^2+binom101^2+binom102^2+cdots+binom1010^2 = ?$$



Can you help me? Thanks in advance.




power-series combinations




share|cite|improve this question













closed as off-topic by asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel Jul 23 at 12:30


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Are you asking for a general solution for $sum_k=0^n binomnk^2$ or only for the special case where $n=10$?
    – mrtaurho
    Jul 23 at 8:08













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











I wanna know the answer of this problem and solve it:



$$binom100^2+binom101^2+binom102^2+cdots+binom1010^2 = ?$$



Can you help me? Thanks in advance.




power-series combinations




share|cite|improve this question













I wanna know the answer of this problem and solve it:



$$binom100^2+binom101^2+binom102^2+cdots+binom1010^2 = ?$$



Can you help me? Thanks in advance.





power-series combinations





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 8:59









Lorenzo B.

1,5402418




1,5402418









asked Jul 23 at 8:03









user3832258

22




22




closed as off-topic by asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel Jul 23 at 12:30


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel Jul 23 at 12:30


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, Martin R, José Carlos Santos, amWhy, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Are you asking for a general solution for $sum_k=0^n binomnk^2$ or only for the special case where $n=10$?
    – mrtaurho
    Jul 23 at 8:08

















  • Are you asking for a general solution for $sum_k=0^n binomnk^2$ or only for the special case where $n=10$?
    – mrtaurho
    Jul 23 at 8:08
















Are you asking for a general solution for $sum_k=0^n binomnk^2$ or only for the special case where $n=10$?
– mrtaurho
Jul 23 at 8:08





Are you asking for a general solution for $sum_k=0^n binomnk^2$ or only for the special case where $n=10$?
– mrtaurho
Jul 23 at 8:08











1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










$$binom2nn = sum_k=0^n binomnk binomnn-k = sum_k=0^n binomnk^2$$
I think it will help






share|cite|improve this answer





















  • I want the Prove of this way. Can you write it?
    – user3832258
    Jul 23 at 8:25










  • Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
    – mrtaurho
    Jul 23 at 8:31










  • Thanks a lot. Can you guide me in proving this math formula?
    – user3832258
    Jul 23 at 8:34






  • 1




    @user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
    – D F
    Jul 23 at 8:34


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










$$binom2nn = sum_k=0^n binomnk binomnn-k = sum_k=0^n binomnk^2$$
I think it will help






share|cite|improve this answer





















  • I want the Prove of this way. Can you write it?
    – user3832258
    Jul 23 at 8:25










  • Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
    – mrtaurho
    Jul 23 at 8:31










  • Thanks a lot. Can you guide me in proving this math formula?
    – user3832258
    Jul 23 at 8:34






  • 1




    @user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
    – D F
    Jul 23 at 8:34















up vote
2
down vote



accepted










$$binom2nn = sum_k=0^n binomnk binomnn-k = sum_k=0^n binomnk^2$$
I think it will help






share|cite|improve this answer





















  • I want the Prove of this way. Can you write it?
    – user3832258
    Jul 23 at 8:25










  • Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
    – mrtaurho
    Jul 23 at 8:31










  • Thanks a lot. Can you guide me in proving this math formula?
    – user3832258
    Jul 23 at 8:34






  • 1




    @user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
    – D F
    Jul 23 at 8:34













up vote
2
down vote



accepted







up vote
2
down vote



accepted






$$binom2nn = sum_k=0^n binomnk binomnn-k = sum_k=0^n binomnk^2$$
I think it will help






share|cite|improve this answer













$$binom2nn = sum_k=0^n binomnk binomnn-k = sum_k=0^n binomnk^2$$
I think it will help







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 23 at 8:09









D F

1,0551218




1,0551218











  • I want the Prove of this way. Can you write it?
    – user3832258
    Jul 23 at 8:25










  • Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
    – mrtaurho
    Jul 23 at 8:31










  • Thanks a lot. Can you guide me in proving this math formula?
    – user3832258
    Jul 23 at 8:34






  • 1




    @user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
    – D F
    Jul 23 at 8:34

















  • I want the Prove of this way. Can you write it?
    – user3832258
    Jul 23 at 8:25










  • Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
    – mrtaurho
    Jul 23 at 8:31










  • Thanks a lot. Can you guide me in proving this math formula?
    – user3832258
    Jul 23 at 8:34






  • 1




    @user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
    – D F
    Jul 23 at 8:34
















I want the Prove of this way. Can you write it?
– user3832258
Jul 23 at 8:25




I want the Prove of this way. Can you write it?
– user3832258
Jul 23 at 8:25












Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
– mrtaurho
Jul 23 at 8:31




Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution.
– mrtaurho
Jul 23 at 8:31












Thanks a lot. Can you guide me in proving this math formula?
– user3832258
Jul 23 at 8:34




Thanks a lot. Can you guide me in proving this math formula?
– user3832258
Jul 23 at 8:34




1




1




@user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
– D F
Jul 23 at 8:34





@user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $ forall k = 0, 1, dots, n$, thus, due to the multiplication rule $binom2nn = sum_k=0^n binomnk binomnn-k$, but $binomnk = binomnn-k$ hence $binom2nn = sum_k=0^n binomnk^2$
– D F
Jul 23 at 8:34



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