Count number of possibilities











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I would like to count the number $N_{p,n}$ of $p$-tuple $(j_1,ldots,j_p)$ such that for some $nle 3p$
$$
j_1+ldots +j_p = n text{ and } 0le j_ile 3 ;;forall iin {1,ldots,p}
$$



I tried some examples:



$N_{1,n}=3$...
but I can't get a general picture






combinatorics







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    up vote
    1
    down vote

    favorite












    I would like to count the number $N_{p,n}$ of $p$-tuple $(j_1,ldots,j_p)$ such that for some $nle 3p$
    $$
    j_1+ldots +j_p = n text{ and } 0le j_ile 3 ;;forall iin {1,ldots,p}
    $$



    I tried some examples:



    $N_{1,n}=3$...
    but I can't get a general picture






    combinatorics







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I would like to count the number $N_{p,n}$ of $p$-tuple $(j_1,ldots,j_p)$ such that for some $nle 3p$
      $$
      j_1+ldots +j_p = n text{ and } 0le j_ile 3 ;;forall iin {1,ldots,p}
      $$



      I tried some examples:



      $N_{1,n}=3$...
      but I can't get a general picture






      combinatorics







      share|cite|improve this question













      I would like to count the number $N_{p,n}$ of $p$-tuple $(j_1,ldots,j_p)$ such that for some $nle 3p$
      $$
      j_1+ldots +j_p = n text{ and } 0le j_ile 3 ;;forall iin {1,ldots,p}
      $$



      I tried some examples:



      $N_{1,n}=3$...
      but I can't get a general picture






      combinatorics




      combinatorics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 at 16:01









      Smilia

      585516




      585516






















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          Number of such $p$-tuples= Coefficient of $x^n$ in $(1+x+x^2+x^3)^p$ with $|x|<1.$



          $(1+x+x^2+x^3)^p=left(frac{1-x^4}{1-x}right)^p=(1-x^4)^p(1-x)^{-p}=(1-x^4)^p(sum_{k=0}^infty(-1)^k {-pchoose k}x^k).$



          So, Number of such $p$-tuples $=sum_{l : 4lleq n} (-1)^l(-1)^{n-4l}{pchoose l}{-pchoose n-4l}.$



          Here I followed the definition ${xchoose k}=frac{x(x-1)ldots(x-k+1)}{k!}.$






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
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            active

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            up vote
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            down vote



            accepted










            Number of such $p$-tuples= Coefficient of $x^n$ in $(1+x+x^2+x^3)^p$ with $|x|<1.$



            $(1+x+x^2+x^3)^p=left(frac{1-x^4}{1-x}right)^p=(1-x^4)^p(1-x)^{-p}=(1-x^4)^p(sum_{k=0}^infty(-1)^k {-pchoose k}x^k).$



            So, Number of such $p$-tuples $=sum_{l : 4lleq n} (-1)^l(-1)^{n-4l}{pchoose l}{-pchoose n-4l}.$



            Here I followed the definition ${xchoose k}=frac{x(x-1)ldots(x-k+1)}{k!}.$






            share|cite|improve this answer

























              up vote
              3
              down vote



              accepted










              Number of such $p$-tuples= Coefficient of $x^n$ in $(1+x+x^2+x^3)^p$ with $|x|<1.$



              $(1+x+x^2+x^3)^p=left(frac{1-x^4}{1-x}right)^p=(1-x^4)^p(1-x)^{-p}=(1-x^4)^p(sum_{k=0}^infty(-1)^k {-pchoose k}x^k).$



              So, Number of such $p$-tuples $=sum_{l : 4lleq n} (-1)^l(-1)^{n-4l}{pchoose l}{-pchoose n-4l}.$



              Here I followed the definition ${xchoose k}=frac{x(x-1)ldots(x-k+1)}{k!}.$






              share|cite|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                Number of such $p$-tuples= Coefficient of $x^n$ in $(1+x+x^2+x^3)^p$ with $|x|<1.$



                $(1+x+x^2+x^3)^p=left(frac{1-x^4}{1-x}right)^p=(1-x^4)^p(1-x)^{-p}=(1-x^4)^p(sum_{k=0}^infty(-1)^k {-pchoose k}x^k).$



                So, Number of such $p$-tuples $=sum_{l : 4lleq n} (-1)^l(-1)^{n-4l}{pchoose l}{-pchoose n-4l}.$



                Here I followed the definition ${xchoose k}=frac{x(x-1)ldots(x-k+1)}{k!}.$






                share|cite|improve this answer












                Number of such $p$-tuples= Coefficient of $x^n$ in $(1+x+x^2+x^3)^p$ with $|x|<1.$



                $(1+x+x^2+x^3)^p=left(frac{1-x^4}{1-x}right)^p=(1-x^4)^p(1-x)^{-p}=(1-x^4)^p(sum_{k=0}^infty(-1)^k {-pchoose k}x^k).$



                So, Number of such $p$-tuples $=sum_{l : 4lleq n} (-1)^l(-1)^{n-4l}{pchoose l}{-pchoose n-4l}.$



                Here I followed the definition ${xchoose k}=frac{x(x-1)ldots(x-k+1)}{k!}.$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 16:19









                John_Wick

                1,134111




                1,134111






























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