Mixing
How many solutions are there to the equation $x + y + z + w = 17$?
$begingroup$
How many solutions are there to the equation $x + y + z + w = 17$ for non-negative integers $w, x, y, z$ ?
I don't know if I'm doing this right, but I guessed that the solution would be $binom{20}{3}$, which equals $1140$. Am I doing this right?
combinatorics discrete-mathematics binomial-coefficients diophantine-equations
asked Dec 15 '11 at 12:30
user754950user754950
175247
$endgroup$
add a comment |
$begingroup$
How many solutions are there to the equation $x + y + z + w = 17$ for non-negative integers $w, x, y, z$ ?
I don't know if I'm doing this right, but I guessed that the solution would be $binom{20}{3}$, which equals $1140$. Am I doing this right?
combinatorics discrete-mathematics binomial-coefficients diophantine-equations
asked Dec 15 '11 at 12:30
user754950user754950
175247
$endgroup$
$begingroup$
Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else?
$endgroup$
– Srivatsan
Dec 15 '11 at 12:33
$begingroup$
They are non-negative integers
$endgroup$
– user754950
Dec 15 '11 at 12:38
1
$begingroup$
See math.stackexchange.com/questions/86820/… and the questions referenced there.
$endgroup$
– Gerry Myerson
Dec 15 '11 at 12:38
add a comment |
$begingroup$
How many solutions are there to the equation $x + y + z + w = 17$ for non-negative integers $w, x, y, z$ ?
I don't know if I'm doing this right, but I guessed that the solution would be $binom{20}{3}$, which equals $1140$. Am I doing this right?
combinatorics discrete-mathematics binomial-coefficients diophantine-equations
asked Dec 15 '11 at 12:30
user754950user754950
175247
$endgroup$
How many solutions are there to the equation $x + y + z + w = 17$ for non-negative integers $w, x, y, z$ ?
I don't know if I'm doing this right, but I guessed that the solution would be $binom{20}{3}$, which equals $1140$. Am I doing this right?
combinatorics discrete-mathematics binomial-coefficients diophantine-equations
combinatorics discrete-mathematics binomial-coefficients diophantine-equations
asked Dec 15 '11 at 12:30
user754950user754950
175247
asked Dec 15 '11 at 12:30
user754950user754950
175247
edited Feb 7 '16 at 10:29
user228113
asked Dec 15 '11 at 12:30
user754950user754950
175247
asked Dec 15 '11 at 12:30
user754950user754950
175247
asked Dec 15 '11 at 12:30
user754950user754950
175247
175247
$begingroup$
Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else?
$endgroup$
– Srivatsan
Dec 15 '11 at 12:33
$begingroup$
They are non-negative integers
$endgroup$
– user754950
Dec 15 '11 at 12:38
1
$begingroup$
See math.stackexchange.com/questions/86820/… and the questions referenced there.
$endgroup$
– Gerry Myerson
Dec 15 '11 at 12:38
add a comment |
$begingroup$
Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else?
$endgroup$
– Srivatsan
Dec 15 '11 at 12:33
$begingroup$
They are non-negative integers
$endgroup$
– user754950
Dec 15 '11 at 12:38
1
$begingroup$
See math.stackexchange.com/questions/86820/… and the questions referenced there.
$endgroup$
– Gerry Myerson
Dec 15 '11 at 12:38
$begingroup$
Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else?
$endgroup$
– Srivatsan
Dec 15 '11 at 12:33
$begingroup$
Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else?
$endgroup$
– Srivatsan
Dec 15 '11 at 12:33
$begingroup$
They are non-negative integers
$endgroup$
– user754950
Dec 15 '11 at 12:38
$begingroup$
They are non-negative integers
$endgroup$
– user754950
Dec 15 '11 at 12:38
1
1
$begingroup$
See math.stackexchange.com/questions/86820/… and the questions referenced there.
$endgroup$
– Gerry Myerson
Dec 15 '11 at 12:38
$begingroup$
See math.stackexchange.com/questions/86820/… and the questions referenced there.
$endgroup$
– Gerry Myerson
Dec 15 '11 at 12:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your result is correct. You are looking for the number of monomials of total degree $17$ in $4$ variables. There are
$$binom{17+4-1}{17}=binom{20}{17}=frac{20cdot 19cdot 18}{3cdot 2} = 20cdot19cdot3 = 1140$$
of these.
A proof would be the following. We associate to any solution $(x,y,z,w)$ a weakly increasing sequence of length $17$, where $0$ appears $x$ times, $1$ appears $y$ times, and so on. For instance, the solution $(4,3,2,8)$ would correspond to
$$(0,0,0,0,1,1,1,2,2,3,3,3,3,3,3,3,3)$$
By adding $i$ to the $i$-th component of that sequence, it becomes strictly increasing. This way, we can see that the solutions correspond to subsets of ${1,ldots,17+4-1}$ of cardinality $17$. Of course, you can replace $17$ and $4$ by $n$ and $d$ to obtain a more general statement.
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
$endgroup$
add a comment |
$begingroup$
Line up $20$ skittles in a row:
+ + + + + + + + + + + + + + + + + + + +
Knock any three of them over:
+ + + x + + + + + + x x + + + + + + + +
Now let $x,y,z,$ and $w$ be the number of skittles left standing between the knocked-over skittles. In this case $x=3, y=6, z=0,$ and $w=8$.
So your answer is correct.
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
$endgroup$
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
1
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your result is correct. You are looking for the number of monomials of total degree $17$ in $4$ variables. There are
$$binom{17+4-1}{17}=binom{20}{17}=frac{20cdot 19cdot 18}{3cdot 2} = 20cdot19cdot3 = 1140$$
of these.
A proof would be the following. We associate to any solution $(x,y,z,w)$ a weakly increasing sequence of length $17$, where $0$ appears $x$ times, $1$ appears $y$ times, and so on. For instance, the solution $(4,3,2,8)$ would correspond to
$$(0,0,0,0,1,1,1,2,2,3,3,3,3,3,3,3,3)$$
By adding $i$ to the $i$-th component of that sequence, it becomes strictly increasing. This way, we can see that the solutions correspond to subsets of ${1,ldots,17+4-1}$ of cardinality $17$. Of course, you can replace $17$ and $4$ by $n$ and $d$ to obtain a more general statement.
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
$endgroup$
add a comment |
$begingroup$
Your result is correct. You are looking for the number of monomials of total degree $17$ in $4$ variables. There are
$$binom{17+4-1}{17}=binom{20}{17}=frac{20cdot 19cdot 18}{3cdot 2} = 20cdot19cdot3 = 1140$$
of these.
A proof would be the following. We associate to any solution $(x,y,z,w)$ a weakly increasing sequence of length $17$, where $0$ appears $x$ times, $1$ appears $y$ times, and so on. For instance, the solution $(4,3,2,8)$ would correspond to
$$(0,0,0,0,1,1,1,2,2,3,3,3,3,3,3,3,3)$$
By adding $i$ to the $i$-th component of that sequence, it becomes strictly increasing. This way, we can see that the solutions correspond to subsets of ${1,ldots,17+4-1}$ of cardinality $17$. Of course, you can replace $17$ and $4$ by $n$ and $d$ to obtain a more general statement.
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
$endgroup$
add a comment |
$begingroup$
Your result is correct. You are looking for the number of monomials of total degree $17$ in $4$ variables. There are
$$binom{17+4-1}{17}=binom{20}{17}=frac{20cdot 19cdot 18}{3cdot 2} = 20cdot19cdot3 = 1140$$
of these.
A proof would be the following. We associate to any solution $(x,y,z,w)$ a weakly increasing sequence of length $17$, where $0$ appears $x$ times, $1$ appears $y$ times, and so on. For instance, the solution $(4,3,2,8)$ would correspond to
$$(0,0,0,0,1,1,1,2,2,3,3,3,3,3,3,3,3)$$
By adding $i$ to the $i$-th component of that sequence, it becomes strictly increasing. This way, we can see that the solutions correspond to subsets of ${1,ldots,17+4-1}$ of cardinality $17$. Of course, you can replace $17$ and $4$ by $n$ and $d$ to obtain a more general statement.
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
$endgroup$
Your result is correct. You are looking for the number of monomials of total degree $17$ in $4$ variables. There are
$$binom{17+4-1}{17}=binom{20}{17}=frac{20cdot 19cdot 18}{3cdot 2} = 20cdot19cdot3 = 1140$$
of these.
A proof would be the following. We associate to any solution $(x,y,z,w)$ a weakly increasing sequence of length $17$, where $0$ appears $x$ times, $1$ appears $y$ times, and so on. For instance, the solution $(4,3,2,8)$ would correspond to
$$(0,0,0,0,1,1,1,2,2,3,3,3,3,3,3,3,3)$$
By adding $i$ to the $i$-th component of that sequence, it becomes strictly increasing. This way, we can see that the solutions correspond to subsets of ${1,ldots,17+4-1}$ of cardinality $17$. Of course, you can replace $17$ and $4$ by $n$ and $d$ to obtain a more general statement.
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
edited Dec 15 '11 at 13:08
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
answered Dec 15 '11 at 12:47
Jesko HüttenhainJesko Hüttenhain
10.6k12357
10.6k12357
add a comment |
add a comment |
$begingroup$
Line up $20$ skittles in a row:
+ + + + + + + + + + + + + + + + + + + +
Knock any three of them over:
+ + + x + + + + + + x x + + + + + + + +
Now let $x,y,z,$ and $w$ be the number of skittles left standing between the knocked-over skittles. In this case $x=3, y=6, z=0,$ and $w=8$.
So your answer is correct.
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
$endgroup$
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
1
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
add a comment |
$begingroup$
Line up $20$ skittles in a row:
+ + + + + + + + + + + + + + + + + + + +
Knock any three of them over:
+ + + x + + + + + + x x + + + + + + + +
Now let $x,y,z,$ and $w$ be the number of skittles left standing between the knocked-over skittles. In this case $x=3, y=6, z=0,$ and $w=8$.
So your answer is correct.
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
$endgroup$
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
1
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
add a comment |
$begingroup$
Line up $20$ skittles in a row:
+ + + + + + + + + + + + + + + + + + + +
Knock any three of them over:
+ + + x + + + + + + x x + + + + + + + +
Now let $x,y,z,$ and $w$ be the number of skittles left standing between the knocked-over skittles. In this case $x=3, y=6, z=0,$ and $w=8$.
So your answer is correct.
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
$endgroup$
Line up $20$ skittles in a row:
+ + + + + + + + + + + + + + + + + + + +
Knock any three of them over:
+ + + x + + + + + + x x + + + + + + + +
Now let $x,y,z,$ and $w$ be the number of skittles left standing between the knocked-over skittles. In this case $x=3, y=6, z=0,$ and $w=8$.
So your answer is correct.
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
answered Dec 15 '11 at 13:19
TonyKTonyK
43k356135
43k356135
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
1
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
add a comment |
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
1
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
$begingroup$
that's beautiful.
$endgroup$
– Jesko Hüttenhain
Dec 15 '11 at 13:22
1
1
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
$begingroup$
I was a bridge programmer a long time ago. In bridge, C + D + H + S = 13.
$endgroup$
– TonyK
Dec 15 '11 at 13:36
add a comment |
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$begingroup$
Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else?
$endgroup$
– Srivatsan
Dec 15 '11 at 12:33
$begingroup$
They are non-negative integers
$endgroup$
– user754950
Dec 15 '11 at 12:38
1
$begingroup$
See math.stackexchange.com/questions/86820/… and the questions referenced there.
$endgroup$
– Gerry Myerson
Dec 15 '11 at 12:38