Summing rolls of three distinct dice and gets $a$, $b$, and $c$. Find the probability of $2a + b + c = 10$.
$begingroup$
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
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
asked Dec 13 '18 at 1:46
user526427user526427
1187
$endgroup$
add a comment |
$begingroup$
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
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
asked Dec 13 '18 at 1:46
user526427user526427
1187
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
asked Dec 13 '18 at 1:46
user526427user526427
1187
$endgroup$
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
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
asked Dec 13 '18 at 1:46
user526427user526427
1187
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
asked Dec 13 '18 at 1:46
user526427user526427
1187
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
edited Dec 13 '18 at 9:19
N. F. Taussig
44.1k93356
44.1k93356
asked Dec 13 '18 at 1:46
user526427user526427
1187
asked Dec 13 '18 at 1:46
user526427user526427
1187
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}
It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
$endgroup$
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.
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
edited Dec 13 '18 at 2:02
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
answered Dec 13 '18 at 1:48
Bram28Bram28
61.9k44793
61.9k44793
add a comment |
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}
It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}
It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}
It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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}
It means $ds{bbx{{14 over 6 times 6 times 6} = {7 over 108}}}$.
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
edited Dec 14 '18 at 18:12
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
answered Dec 13 '18 at 2:37
Felix MarinFelix Marin
67.8k7107142
67.8k7107142
add a comment |
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
$endgroup$
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.
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
answered Dec 13 '18 at 2:40
T. FoT. Fo
466311
466311
add a comment |
add a comment |
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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.
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– hardmath
Dec 13 '18 at 1:51
$begingroup$
You should get $7+5+3 +1 = 16$.
$endgroup$
– rsadhvika
Dec 13 '18 at 2:03