Index summation over a binomial coefficiant series [closed]












-2












$begingroup$


Can the following sum be reduced to an analytic expression?



$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$






summation binomial-coefficients







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closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52


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." – callculus, NCh, Andrew, max_zorn, saz

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
    $endgroup$
    – callculus
    Jan 27 at 16:20








  • 2




    $begingroup$
    This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
    $endgroup$
    – user376343
    Jan 27 at 16:24










  • $begingroup$
    For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
    $endgroup$
    – Ethan Bolker
    Jan 27 at 16:37
















-2












$begingroup$


Can the following sum be reduced to an analytic expression?



$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$






summation binomial-coefficients







share|cite|improve this question









$endgroup$



closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52


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." – callculus, NCh, Andrew, max_zorn, saz

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
    $endgroup$
    – callculus
    Jan 27 at 16:20








  • 2




    $begingroup$
    This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
    $endgroup$
    – user376343
    Jan 27 at 16:24










  • $begingroup$
    For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
    $endgroup$
    – Ethan Bolker
    Jan 27 at 16:37














-2












-2








-2





$begingroup$


Can the following sum be reduced to an analytic expression?



$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$






summation binomial-coefficients







share|cite|improve this question









$endgroup$




Can the following sum be reduced to an analytic expression?



$$sum_{m=0}^Nfrac{N!}{m!(N-m)!}$$






summation binomial-coefficients




summation binomial-coefficients






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 27 at 16:14









jarheadjarhead

1308




1308




closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52


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." – callculus, NCh, Andrew, max_zorn, saz

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52


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." – callculus, NCh, Andrew, max_zorn, saz

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
    $endgroup$
    – callculus
    Jan 27 at 16:20








  • 2




    $begingroup$
    This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
    $endgroup$
    – user376343
    Jan 27 at 16:24










  • $begingroup$
    For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
    $endgroup$
    – Ethan Bolker
    Jan 27 at 16:37














  • 1




    $begingroup$
    Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
    $endgroup$
    – callculus
    Jan 27 at 16:20








  • 2




    $begingroup$
    This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
    $endgroup$
    – user376343
    Jan 27 at 16:24










  • $begingroup$
    For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
    $endgroup$
    – Ethan Bolker
    Jan 27 at 16:37








1




1




$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20






$begingroup$
Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear.
$endgroup$
– callculus
Jan 27 at 16:20






2




2




$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24




$begingroup$
This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$
$endgroup$
– user376343
Jan 27 at 16:24












$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37




$begingroup$
For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it.
$endgroup$
– Ethan Bolker
Jan 27 at 16:37










1 Answer
1






active

oldest

votes


















1












$begingroup$

Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$



Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$



    Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$



      Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$



        Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$






        share|cite|improve this answer









        $endgroup$



        Note: $$(1+x)^n= ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+ldots +^nC_n.1^0.x^n$$



        Put x = 1, to recover $$sum_{m=0}^{n} {^n}C_m = 2^n$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 at 16:32









        Exp ikxExp ikx

        4489




        4489















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