Binomial Theorem with Three Terms











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$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.










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  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42















up vote
4
down vote

favorite
1












$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.










share|cite|improve this question


















  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.










share|cite|improve this question













$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.







combinatorics binomial-coefficients






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asked Nov 19 at 10:18









ultralight

396




396








  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42














  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42








1




1




Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
Nov 19 at 12:16




Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
Nov 19 at 12:16




1




1




A "binomial with three terms" should probably be called a trinomial.
– Torsten Schoeneberg
Nov 20 at 8:42




A "binomial with three terms" should probably be called a trinomial.
– Torsten Schoeneberg
Nov 20 at 8:42










5 Answers
5






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up vote
13
down vote



accepted










In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



Thus the coefficient of $x^8$ is $8(35)+21 = 301$






share|cite|improve this answer





















  • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
    – mathnoob
    Nov 19 at 10:47










  • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
    – Mohammad Riazi-Kermani
    Nov 19 at 10:51


















up vote
11
down vote













The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$



where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$

Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





  • $i=4$, $k=0$, $j=3$;


  • $i=5$, $k=2$, $j=0$.


Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$






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    up vote
    5
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    Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
    Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



    Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



    If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



    If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



    If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



    so the answer is $301$.






    share|cite|improve this answer






























      up vote
      2
      down vote













      The answer is 301.



      Just trust your plan of the twofold use of the binomial formula:



      First step



      $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



      Second step



      $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



      Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



      $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



      Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






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        To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



        This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



        7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

        7+4+4, 4+7+4, 4+4+7

        6+6+3, 6+3+6, 3+6+6

        6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



        That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






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        • What about to undelete that OP?
          – gimusi
          Nov 26 at 9:33











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        5 Answers
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        5 Answers
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        up vote
        13
        down vote



        accepted










        In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



        There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



        Thus the coefficient of $x^8$ is $8(35)+21 = 301$






        share|cite|improve this answer





















        • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
          – mathnoob
          Nov 19 at 10:47










        • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
          – Mohammad Riazi-Kermani
          Nov 19 at 10:51















        up vote
        13
        down vote



        accepted










        In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



        There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



        Thus the coefficient of $x^8$ is $8(35)+21 = 301$






        share|cite|improve this answer





















        • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
          – mathnoob
          Nov 19 at 10:47










        • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
          – Mohammad Riazi-Kermani
          Nov 19 at 10:51













        up vote
        13
        down vote



        accepted







        up vote
        13
        down vote



        accepted






        In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



        There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



        Thus the coefficient of $x^8$ is $8(35)+21 = 301$






        share|cite|improve this answer












        In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



        There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



        Thus the coefficient of $x^8$ is $8(35)+21 = 301$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 10:44









        Mohammad Riazi-Kermani

        40.3k41958




        40.3k41958












        • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
          – mathnoob
          Nov 19 at 10:47










        • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
          – Mohammad Riazi-Kermani
          Nov 19 at 10:51


















        • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
          – mathnoob
          Nov 19 at 10:47










        • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
          – Mohammad Riazi-Kermani
          Nov 19 at 10:51
















        But in the first type, don't you also have to multiply by $(x^{-1})^3$?
        – mathnoob
        Nov 19 at 10:47




        But in the first type, don't you also have to multiply by $(x^{-1})^3$?
        – mathnoob
        Nov 19 at 10:47












        @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
        – Mohammad Riazi-Kermani
        Nov 19 at 10:51




        @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
        – Mohammad Riazi-Kermani
        Nov 19 at 10:51










        up vote
        11
        down vote













        The multinomial theorem can come to the rescue:
        $$
        (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
        $$



        where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



        Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
        $$
        2i-k=8,qquad i+kle 7
        $$

        Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





        • $i=4$, $k=0$, $j=3$;


        • $i=5$, $k=2$, $j=0$.


        Thus the coefficient is
        $$
        2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
        8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
        $$






        share|cite|improve this answer



























          up vote
          11
          down vote













          The multinomial theorem can come to the rescue:
          $$
          (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
          $$



          where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



          Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
          $$
          2i-k=8,qquad i+kle 7
          $$

          Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





          • $i=4$, $k=0$, $j=3$;


          • $i=5$, $k=2$, $j=0$.


          Thus the coefficient is
          $$
          2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
          8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
          $$






          share|cite|improve this answer

























            up vote
            11
            down vote










            up vote
            11
            down vote









            The multinomial theorem can come to the rescue:
            $$
            (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
            $$



            where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



            Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
            $$
            2i-k=8,qquad i+kle 7
            $$

            Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





            • $i=4$, $k=0$, $j=3$;


            • $i=5$, $k=2$, $j=0$.


            Thus the coefficient is
            $$
            2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
            8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
            $$






            share|cite|improve this answer














            The multinomial theorem can come to the rescue:
            $$
            (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
            $$



            where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



            Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
            $$
            2i-k=8,qquad i+kle 7
            $$

            Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





            • $i=4$, $k=0$, $j=3$;


            • $i=5$, $k=2$, $j=0$.


            Thus the coefficient is
            $$
            2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
            8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 19 at 21:13

























            answered Nov 19 at 15:26









            egreg

            176k1384198




            176k1384198






















                up vote
                5
                down vote













                Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                so the answer is $301$.






                share|cite|improve this answer



























                  up vote
                  5
                  down vote













                  Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                  Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                  Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                  If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                  If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                  If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                  so the answer is $301$.






                  share|cite|improve this answer

























                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









                    Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                    Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                    Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                    If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                    If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                    If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                    so the answer is $301$.






                    share|cite|improve this answer














                    Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                    Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                    Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                    If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                    If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                    If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                    so the answer is $301$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 19 at 15:37









                    Rad80

                    30718




                    30718










                    answered Nov 19 at 10:42









                    greedoid

                    36.4k114591




                    36.4k114591






















                        up vote
                        2
                        down vote













                        The answer is 301.



                        Just trust your plan of the twofold use of the binomial formula:



                        First step



                        $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                        Second step



                        $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                        Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                        $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                        Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






                        share|cite|improve this answer



























                          up vote
                          2
                          down vote













                          The answer is 301.



                          Just trust your plan of the twofold use of the binomial formula:



                          First step



                          $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                          Second step



                          $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                          Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                          $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                          Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






                          share|cite|improve this answer

























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            The answer is 301.



                            Just trust your plan of the twofold use of the binomial formula:



                            First step



                            $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                            Second step



                            $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                            Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                            $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                            Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






                            share|cite|improve this answer














                            The answer is 301.



                            Just trust your plan of the twofold use of the binomial formula:



                            First step



                            $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                            Second step



                            $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                            Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                            $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                            Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 19 at 11:29

























                            answered Nov 19 at 11:09









                            Dr. Wolfgang Hintze

                            3,035616




                            3,035616






















                                up vote
                                1
                                down vote













                                To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                7+4+4, 4+7+4, 4+4+7

                                6+6+3, 6+3+6, 3+6+6

                                6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






                                share|cite|improve this answer





















                                • What about to undelete that OP?
                                  – gimusi
                                  Nov 26 at 9:33















                                up vote
                                1
                                down vote













                                To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                7+4+4, 4+7+4, 4+4+7

                                6+6+3, 6+3+6, 3+6+6

                                6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






                                share|cite|improve this answer





















                                • What about to undelete that OP?
                                  – gimusi
                                  Nov 26 at 9:33













                                up vote
                                1
                                down vote










                                up vote
                                1
                                down vote









                                To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                7+4+4, 4+7+4, 4+4+7

                                6+6+3, 6+3+6, 3+6+6

                                6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






                                share|cite|improve this answer












                                To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                7+4+4, 4+7+4, 4+4+7

                                6+6+3, 6+3+6, 3+6+6

                                6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 19 at 18:37









                                Acccumulation

                                6,5812616




                                6,5812616












                                • What about to undelete that OP?
                                  – gimusi
                                  Nov 26 at 9:33


















                                • What about to undelete that OP?
                                  – gimusi
                                  Nov 26 at 9:33
















                                What about to undelete that OP?
                                – gimusi
                                Nov 26 at 9:33




                                What about to undelete that OP?
                                – gimusi
                                Nov 26 at 9:33


















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