Summing rolls of three distinct dice and gets $a$, $b$, and $c$. Find the probability of $2a + b + c = 10$.












0












$begingroup$



Summing rolls of three distinct dice and gets $a$, $b$, and $c$. Find the probability of $2a + b + c = 10$.




My answer:



a b c

4 1 1

3 2 2

2 3 3

1 4 4



3 3 1

3 1 3

2 1 5

2 5 1



2 4 2

2 2 4

1 5 3

1 3 5



Total $12$ cases out of $6^3 implies $ probability is $1/18$. Is this correct? Please check.






probability combinatorics







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$endgroup$








  • 1




    $begingroup$
    It's poor practice to use the title exclusively to give a problem statement. Indeed the full problem statement should appear in the body of the Question, regardless of whether the title gives a terse version of the problem.
    $endgroup$
    – hardmath
    Dec 13 '18 at 1:51










  • $begingroup$
    You should get $7+5+3 +1 = 16$.
    $endgroup$
    – rsadhvika
    Dec 13 '18 at 2:03
















0












$begingroup$



Summing rolls of three distinct dice and gets $a$, $b$, and $c$. Find the probability of $2a + b + c = 10$.




My answer:



a b c

4 1 1

3 2 2

2 3 3

1 4 4



3 3 1

3 1 3

2 1 5

2 5 1



2 4 2

2 2 4

1 5 3

1 3 5



Total $12$ cases out of $6^3 implies $ probability is $1/18$. Is this correct? Please check.






probability combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's poor practice to use the title exclusively to give a problem statement. Indeed the full problem statement should appear in the body of the Question, regardless of whether the title gives a terse version of the problem.
    $endgroup$
    – hardmath
    Dec 13 '18 at 1:51










  • $begingroup$
    You should get $7+5+3 +1 = 16$.
    $endgroup$
    – rsadhvika
    Dec 13 '18 at 2:03














0












0








0





$begingroup$



Summing rolls of three distinct dice and gets $a$, $b$, and $c$. Find the probability of $2a + b + c = 10$.




My answer:



a b c

4 1 1

3 2 2

2 3 3

1 4 4



3 3 1

3 1 3

2 1 5

2 5 1



2 4 2

2 2 4

1 5 3

1 3 5



Total $12$ cases out of $6^3 implies $ probability is $1/18$. Is this correct? Please check.






probability combinatorics







share|cite|improve this question











$endgroup$





Summing rolls of three distinct dice and gets $a$, $b$, and $c$. Find the probability of $2a + b + c = 10$.




My answer:



a b c

4 1 1

3 2 2

2 3 3

1 4 4



3 3 1

3 1 3

2 1 5

2 5 1



2 4 2

2 2 4

1 5 3

1 3 5



Total $12$ cases out of $6^3 implies $ probability is $1/18$. Is this correct? Please check.






probability combinatorics




probability combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 9:19









N. F. Taussig

44.5k103357




44.5k103357










asked Dec 13 '18 at 1:46









user526427user526427

1187




1187








  • 1




    $begingroup$
    It's poor practice to use the title exclusively to give a problem statement. Indeed the full problem statement should appear in the body of the Question, regardless of whether the title gives a terse version of the problem.
    $endgroup$
    – hardmath
    Dec 13 '18 at 1:51










  • $begingroup$
    You should get $7+5+3 +1 = 16$.
    $endgroup$
    – rsadhvika
    Dec 13 '18 at 2:03














  • 1




    $begingroup$
    It's poor practice to use the title exclusively to give a problem statement. Indeed the full problem statement should appear in the body of the Question, regardless of whether the title gives a terse version of the problem.
    $endgroup$
    – hardmath
    Dec 13 '18 at 1:51










  • $begingroup$
    You should get $7+5+3 +1 = 16$.
    $endgroup$
    – rsadhvika
    Dec 13 '18 at 2:03








1




1




$begingroup$
It's poor practice to use the title exclusively to give a problem statement. Indeed the full problem statement should appear in the body of the Question, regardless of whether the title gives a terse version of the problem.
$endgroup$
– hardmath
Dec 13 '18 at 1:51




$begingroup$
It's poor practice to use the title exclusively to give a problem statement. Indeed the full problem statement should appear in the body of the Question, regardless of whether the title gives a terse version of the problem.
$endgroup$
– hardmath
Dec 13 '18 at 1:51












$begingroup$
You should get $7+5+3 +1 = 16$.
$endgroup$
– rsadhvika
Dec 13 '18 at 2:03




$begingroup$
You should get $7+5+3 +1 = 16$.
$endgroup$
– rsadhvika
Dec 13 '18 at 2:03










3 Answers
3






active

oldest

votes


















3












$begingroup$

No. You forgot $1,2,6$ and $1,6,2$



I think you would not have missed those had you generated the combination a little more systematically. That is, starting with $a=4$ ( which is clearly the maximum for $a$), you indeed get only one option: $4,1,1$. OK, so let's move to $a=3$. Now let's figure out all options with that, starting with the highest possible value of $b$: $3,3,1$, $3,2,2$, $3,1,3$, and that's it. And only now go down to $a=2$, etc. Indeed, doing this, I immediately spotted the missing two combinations.



Looking at your combinations, there is a certain systematicity to it ... but not enough.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    begin{align}
    &bbox[10px,#ffd]{sum_{a = 1}^{6}{1 over 6}
    sum_{b = 1}^{6}{1 over 6}sum_{c = 1}^{6}
    {1 over 6}bracks{z^{10}}z^{2a + b + c}}
    \[5mm] = &
    {1 over 216}bracks{z^{10}}
    bracks{sum_{a = 1}^{6}pars{z^{2}}^{a}}
    pars{sum_{b = 1}^{6}z^{b}}pars{sum_{b = 1}^{6}z^{c}}
    \[5mm] = &
    {1 over 216}bracks{z^{10}}
    pars{z^{2},{z^{12} - 1 over z^{2} - 1}}
    pars{z,{z^{6} - 1 over z - 1}}
    pars{z,{z^{6} - 1 over z - 1}}
    \[5mm] = &
    {1 over 216}bracks{z^{6}}{-z^{24} + 2z^{18} - 2z^{6} + 1 over
    pars{1 - z^{2}}pars{1 - z}^{2}} =
    {1 over 216}bracks{z^{6}}{-2z^{6} + 1 over
    pars{1 - z^{2}}pars{1 - z}^{2}}
    \[5mm] = &
    -,{1 over 108} + {1 over 216},bracks{z^{6}}
    sum_{m = 0}^{infty}z^{2m}
    sum_{n = 0}^{infty}{-2 choose n}pars{-z}^{n}
    \[5mm] = &
    -,{1 over 108} + {1 over 216}
    sum_{m = 0}^{infty}
    sum_{n = 0}^{infty}bracks{{2 + n - 1choose n}
    pars{-1}^{n}}pars{-1}^{n}bracks{2m + n = 6}
    \[5mm] = &
    -,{1 over 108} + {1 over 216}
    sum_{m = 0}^{infty}sum_{n = 0}^{infty}pars{n + 1}
    bracks{n = 6 - 2m}
    \[5mm] = &
    -,{1 over 108} + {1 over 216}
    sum_{m = 0}^{infty}pars{7 - 2m}bracks{6 - 2m geq 0}
    \[5mm] = &
    -,{1 over 108} + {1 over 216}
    sum_{m = 0}^{3}pars{7 - 2m} =
    bbx{7 over 108} approx 0.0648
    end{align}







    enter image description here



    It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Hint: By parity, you know that b and c are both even or both odd.



      From here, consider the stars and bars method of counting (multichoose). There will be an edge case or two, but this will simplify your job significantly.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        No. You forgot $1,2,6$ and $1,6,2$



        I think you would not have missed those had you generated the combination a little more systematically. That is, starting with $a=4$ ( which is clearly the maximum for $a$), you indeed get only one option: $4,1,1$. OK, so let's move to $a=3$. Now let's figure out all options with that, starting with the highest possible value of $b$: $3,3,1$, $3,2,2$, $3,1,3$, and that's it. And only now go down to $a=2$, etc. Indeed, doing this, I immediately spotted the missing two combinations.



        Looking at your combinations, there is a certain systematicity to it ... but not enough.






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          No. You forgot $1,2,6$ and $1,6,2$



          I think you would not have missed those had you generated the combination a little more systematically. That is, starting with $a=4$ ( which is clearly the maximum for $a$), you indeed get only one option: $4,1,1$. OK, so let's move to $a=3$. Now let's figure out all options with that, starting with the highest possible value of $b$: $3,3,1$, $3,2,2$, $3,1,3$, and that's it. And only now go down to $a=2$, etc. Indeed, doing this, I immediately spotted the missing two combinations.



          Looking at your combinations, there is a certain systematicity to it ... but not enough.






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            No. You forgot $1,2,6$ and $1,6,2$



            I think you would not have missed those had you generated the combination a little more systematically. That is, starting with $a=4$ ( which is clearly the maximum for $a$), you indeed get only one option: $4,1,1$. OK, so let's move to $a=3$. Now let's figure out all options with that, starting with the highest possible value of $b$: $3,3,1$, $3,2,2$, $3,1,3$, and that's it. And only now go down to $a=2$, etc. Indeed, doing this, I immediately spotted the missing two combinations.



            Looking at your combinations, there is a certain systematicity to it ... but not enough.






            share|cite|improve this answer











            $endgroup$



            No. You forgot $1,2,6$ and $1,6,2$



            I think you would not have missed those had you generated the combination a little more systematically. That is, starting with $a=4$ ( which is clearly the maximum for $a$), you indeed get only one option: $4,1,1$. OK, so let's move to $a=3$. Now let's figure out all options with that, starting with the highest possible value of $b$: $3,3,1$, $3,2,2$, $3,1,3$, and that's it. And only now go down to $a=2$, etc. Indeed, doing this, I immediately spotted the missing two combinations.



            Looking at your combinations, there is a certain systematicity to it ... but not enough.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 13 '18 at 2:02

























            answered Dec 13 '18 at 1:48









            Bram28Bram28

            63.2k44793




            63.2k44793























                1












                $begingroup$

                $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                newcommand{dd}{mathrm{d}}
                newcommand{ds}[1]{displaystyle{#1}}
                newcommand{expo}[1]{,mathrm{e}^{#1},}
                newcommand{ic}{mathrm{i}}
                newcommand{mc}[1]{mathcal{#1}}
                newcommand{mrm}[1]{mathrm{#1}}
                newcommand{pars}[1]{left(,{#1},right)}
                newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                newcommand{verts}[1]{leftvert,{#1},rightvert}$
                begin{align}
                &bbox[10px,#ffd]{sum_{a = 1}^{6}{1 over 6}
                sum_{b = 1}^{6}{1 over 6}sum_{c = 1}^{6}
                {1 over 6}bracks{z^{10}}z^{2a + b + c}}
                \[5mm] = &
                {1 over 216}bracks{z^{10}}
                bracks{sum_{a = 1}^{6}pars{z^{2}}^{a}}
                pars{sum_{b = 1}^{6}z^{b}}pars{sum_{b = 1}^{6}z^{c}}
                \[5mm] = &
                {1 over 216}bracks{z^{10}}
                pars{z^{2},{z^{12} - 1 over z^{2} - 1}}
                pars{z,{z^{6} - 1 over z - 1}}
                pars{z,{z^{6} - 1 over z - 1}}
                \[5mm] = &
                {1 over 216}bracks{z^{6}}{-z^{24} + 2z^{18} - 2z^{6} + 1 over
                pars{1 - z^{2}}pars{1 - z}^{2}} =
                {1 over 216}bracks{z^{6}}{-2z^{6} + 1 over
                pars{1 - z^{2}}pars{1 - z}^{2}}
                \[5mm] = &
                -,{1 over 108} + {1 over 216},bracks{z^{6}}
                sum_{m = 0}^{infty}z^{2m}
                sum_{n = 0}^{infty}{-2 choose n}pars{-z}^{n}
                \[5mm] = &
                -,{1 over 108} + {1 over 216}
                sum_{m = 0}^{infty}
                sum_{n = 0}^{infty}bracks{{2 + n - 1choose n}
                pars{-1}^{n}}pars{-1}^{n}bracks{2m + n = 6}
                \[5mm] = &
                -,{1 over 108} + {1 over 216}
                sum_{m = 0}^{infty}sum_{n = 0}^{infty}pars{n + 1}
                bracks{n = 6 - 2m}
                \[5mm] = &
                -,{1 over 108} + {1 over 216}
                sum_{m = 0}^{infty}pars{7 - 2m}bracks{6 - 2m geq 0}
                \[5mm] = &
                -,{1 over 108} + {1 over 216}
                sum_{m = 0}^{3}pars{7 - 2m} =
                bbx{7 over 108} approx 0.0648
                end{align}







                enter image description here



                It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                  newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                  newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                  newcommand{dd}{mathrm{d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,mathrm{e}^{#1},}
                  newcommand{ic}{mathrm{i}}
                  newcommand{mc}[1]{mathcal{#1}}
                  newcommand{mrm}[1]{mathrm{#1}}
                  newcommand{pars}[1]{left(,{#1},right)}
                  newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                  newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert,{#1},rightvert}$
                  begin{align}
                  &bbox[10px,#ffd]{sum_{a = 1}^{6}{1 over 6}
                  sum_{b = 1}^{6}{1 over 6}sum_{c = 1}^{6}
                  {1 over 6}bracks{z^{10}}z^{2a + b + c}}
                  \[5mm] = &
                  {1 over 216}bracks{z^{10}}
                  bracks{sum_{a = 1}^{6}pars{z^{2}}^{a}}
                  pars{sum_{b = 1}^{6}z^{b}}pars{sum_{b = 1}^{6}z^{c}}
                  \[5mm] = &
                  {1 over 216}bracks{z^{10}}
                  pars{z^{2},{z^{12} - 1 over z^{2} - 1}}
                  pars{z,{z^{6} - 1 over z - 1}}
                  pars{z,{z^{6} - 1 over z - 1}}
                  \[5mm] = &
                  {1 over 216}bracks{z^{6}}{-z^{24} + 2z^{18} - 2z^{6} + 1 over
                  pars{1 - z^{2}}pars{1 - z}^{2}} =
                  {1 over 216}bracks{z^{6}}{-2z^{6} + 1 over
                  pars{1 - z^{2}}pars{1 - z}^{2}}
                  \[5mm] = &
                  -,{1 over 108} + {1 over 216},bracks{z^{6}}
                  sum_{m = 0}^{infty}z^{2m}
                  sum_{n = 0}^{infty}{-2 choose n}pars{-z}^{n}
                  \[5mm] = &
                  -,{1 over 108} + {1 over 216}
                  sum_{m = 0}^{infty}
                  sum_{n = 0}^{infty}bracks{{2 + n - 1choose n}
                  pars{-1}^{n}}pars{-1}^{n}bracks{2m + n = 6}
                  \[5mm] = &
                  -,{1 over 108} + {1 over 216}
                  sum_{m = 0}^{infty}sum_{n = 0}^{infty}pars{n + 1}
                  bracks{n = 6 - 2m}
                  \[5mm] = &
                  -,{1 over 108} + {1 over 216}
                  sum_{m = 0}^{infty}pars{7 - 2m}bracks{6 - 2m geq 0}
                  \[5mm] = &
                  -,{1 over 108} + {1 over 216}
                  sum_{m = 0}^{3}pars{7 - 2m} =
                  bbx{7 over 108} approx 0.0648
                  end{align}







                  enter image description here



                  It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                    newcommand{dd}{mathrm{d}}
                    newcommand{ds}[1]{displaystyle{#1}}
                    newcommand{expo}[1]{,mathrm{e}^{#1},}
                    newcommand{ic}{mathrm{i}}
                    newcommand{mc}[1]{mathcal{#1}}
                    newcommand{mrm}[1]{mathrm{#1}}
                    newcommand{pars}[1]{left(,{#1},right)}
                    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                    newcommand{verts}[1]{leftvert,{#1},rightvert}$
                    begin{align}
                    &bbox[10px,#ffd]{sum_{a = 1}^{6}{1 over 6}
                    sum_{b = 1}^{6}{1 over 6}sum_{c = 1}^{6}
                    {1 over 6}bracks{z^{10}}z^{2a + b + c}}
                    \[5mm] = &
                    {1 over 216}bracks{z^{10}}
                    bracks{sum_{a = 1}^{6}pars{z^{2}}^{a}}
                    pars{sum_{b = 1}^{6}z^{b}}pars{sum_{b = 1}^{6}z^{c}}
                    \[5mm] = &
                    {1 over 216}bracks{z^{10}}
                    pars{z^{2},{z^{12} - 1 over z^{2} - 1}}
                    pars{z,{z^{6} - 1 over z - 1}}
                    pars{z,{z^{6} - 1 over z - 1}}
                    \[5mm] = &
                    {1 over 216}bracks{z^{6}}{-z^{24} + 2z^{18} - 2z^{6} + 1 over
                    pars{1 - z^{2}}pars{1 - z}^{2}} =
                    {1 over 216}bracks{z^{6}}{-2z^{6} + 1 over
                    pars{1 - z^{2}}pars{1 - z}^{2}}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216},bracks{z^{6}}
                    sum_{m = 0}^{infty}z^{2m}
                    sum_{n = 0}^{infty}{-2 choose n}pars{-z}^{n}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{infty}
                    sum_{n = 0}^{infty}bracks{{2 + n - 1choose n}
                    pars{-1}^{n}}pars{-1}^{n}bracks{2m + n = 6}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{infty}sum_{n = 0}^{infty}pars{n + 1}
                    bracks{n = 6 - 2m}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{infty}pars{7 - 2m}bracks{6 - 2m geq 0}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{3}pars{7 - 2m} =
                    bbx{7 over 108} approx 0.0648
                    end{align}







                    enter image description here



                    It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.






                    share|cite|improve this answer











                    $endgroup$



                    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                    newcommand{dd}{mathrm{d}}
                    newcommand{ds}[1]{displaystyle{#1}}
                    newcommand{expo}[1]{,mathrm{e}^{#1},}
                    newcommand{ic}{mathrm{i}}
                    newcommand{mc}[1]{mathcal{#1}}
                    newcommand{mrm}[1]{mathrm{#1}}
                    newcommand{pars}[1]{left(,{#1},right)}
                    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                    newcommand{verts}[1]{leftvert,{#1},rightvert}$
                    begin{align}
                    &bbox[10px,#ffd]{sum_{a = 1}^{6}{1 over 6}
                    sum_{b = 1}^{6}{1 over 6}sum_{c = 1}^{6}
                    {1 over 6}bracks{z^{10}}z^{2a + b + c}}
                    \[5mm] = &
                    {1 over 216}bracks{z^{10}}
                    bracks{sum_{a = 1}^{6}pars{z^{2}}^{a}}
                    pars{sum_{b = 1}^{6}z^{b}}pars{sum_{b = 1}^{6}z^{c}}
                    \[5mm] = &
                    {1 over 216}bracks{z^{10}}
                    pars{z^{2},{z^{12} - 1 over z^{2} - 1}}
                    pars{z,{z^{6} - 1 over z - 1}}
                    pars{z,{z^{6} - 1 over z - 1}}
                    \[5mm] = &
                    {1 over 216}bracks{z^{6}}{-z^{24} + 2z^{18} - 2z^{6} + 1 over
                    pars{1 - z^{2}}pars{1 - z}^{2}} =
                    {1 over 216}bracks{z^{6}}{-2z^{6} + 1 over
                    pars{1 - z^{2}}pars{1 - z}^{2}}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216},bracks{z^{6}}
                    sum_{m = 0}^{infty}z^{2m}
                    sum_{n = 0}^{infty}{-2 choose n}pars{-z}^{n}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{infty}
                    sum_{n = 0}^{infty}bracks{{2 + n - 1choose n}
                    pars{-1}^{n}}pars{-1}^{n}bracks{2m + n = 6}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{infty}sum_{n = 0}^{infty}pars{n + 1}
                    bracks{n = 6 - 2m}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{infty}pars{7 - 2m}bracks{6 - 2m geq 0}
                    \[5mm] = &
                    -,{1 over 108} + {1 over 216}
                    sum_{m = 0}^{3}pars{7 - 2m} =
                    bbx{7 over 108} approx 0.0648
                    end{align}







                    enter image description here



                    It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.







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                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 14 '18 at 18:12

























                    answered Dec 13 '18 at 2:37









                    Felix MarinFelix Marin

                    68.2k7109143




                    68.2k7109143























                        0












                        $begingroup$

                        Hint: By parity, you know that b and c are both even or both odd.



                        From here, consider the stars and bars method of counting (multichoose). There will be an edge case or two, but this will simplify your job significantly.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Hint: By parity, you know that b and c are both even or both odd.



                          From here, consider the stars and bars method of counting (multichoose). There will be an edge case or two, but this will simplify your job significantly.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Hint: By parity, you know that b and c are both even or both odd.



                            From here, consider the stars and bars method of counting (multichoose). There will be an edge case or two, but this will simplify your job significantly.






                            share|cite|improve this answer









                            $endgroup$



                            Hint: By parity, you know that b and c are both even or both odd.



                            From here, consider the stars and bars method of counting (multichoose). There will be an edge case or two, but this will simplify your job significantly.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 13 '18 at 2:40









                            T. FoT. Fo

                            466311




                            466311






























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