ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$begingroup$
[1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$bf{My; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,zin mathbb{Z^{+}}$
$bullet $ If $x=1$, Then $y+z=9$, So $(y,z) = (2,7);,(3,6);,(4,5);(5,4);,(6,3);,(7,2)$
$bullet $ If $x=2$ Then $y+z=8$, So $(y,z) = (1,7);,(3,5);,(5,3);(7,1)$
$bullet $ If $x=3$ Then $y+z=7$, So $(y,z) = (1,6);,(2,5);,(5,2);(6,1)$
$bullet $ If $x=4$ Then $y+z=6$, So $(y,z) = (1,5);,(5,1)$.
$bullet $ If $x=5$ Then $y+z=5$, So $(y,z) = (1,4);,(2,3);,(3,2);(4,1)$
$bullet $ If $x=6$ Then $y+z=4$, So $(y,z) = (1,3);,(3,1)$
$bullet $ If $x=7$ Then $y+z=3$, So $(y,z) = (1,2);,(2,1)$
So Total unordered pair is $ = 24$
My Question is , is there is any other Method to calculate the ordered pair in less complex way
because above is very Lengthy method
Help Required
Thanks.
algebra-precalculus number-theory
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
$endgroup$
add a comment |
$begingroup$
[1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$bf{My; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,zin mathbb{Z^{+}}$
$bullet $ If $x=1$, Then $y+z=9$, So $(y,z) = (2,7);,(3,6);,(4,5);(5,4);,(6,3);,(7,2)$
$bullet $ If $x=2$ Then $y+z=8$, So $(y,z) = (1,7);,(3,5);,(5,3);(7,1)$
$bullet $ If $x=3$ Then $y+z=7$, So $(y,z) = (1,6);,(2,5);,(5,2);(6,1)$
$bullet $ If $x=4$ Then $y+z=6$, So $(y,z) = (1,5);,(5,1)$.
$bullet $ If $x=5$ Then $y+z=5$, So $(y,z) = (1,4);,(2,3);,(3,2);(4,1)$
$bullet $ If $x=6$ Then $y+z=4$, So $(y,z) = (1,3);,(3,1)$
$bullet $ If $x=7$ Then $y+z=3$, So $(y,z) = (1,2);,(2,1)$
So Total unordered pair is $ = 24$
My Question is , is there is any other Method to calculate the ordered pair in less complex way
because above is very Lengthy method
Help Required
Thanks.
algebra-precalculus number-theory
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
$endgroup$
add a comment |
$begingroup$
[1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$bf{My; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,zin mathbb{Z^{+}}$
$bullet $ If $x=1$, Then $y+z=9$, So $(y,z) = (2,7);,(3,6);,(4,5);(5,4);,(6,3);,(7,2)$
$bullet $ If $x=2$ Then $y+z=8$, So $(y,z) = (1,7);,(3,5);,(5,3);(7,1)$
$bullet $ If $x=3$ Then $y+z=7$, So $(y,z) = (1,6);,(2,5);,(5,2);(6,1)$
$bullet $ If $x=4$ Then $y+z=6$, So $(y,z) = (1,5);,(5,1)$.
$bullet $ If $x=5$ Then $y+z=5$, So $(y,z) = (1,4);,(2,3);,(3,2);(4,1)$
$bullet $ If $x=6$ Then $y+z=4$, So $(y,z) = (1,3);,(3,1)$
$bullet $ If $x=7$ Then $y+z=3$, So $(y,z) = (1,2);,(2,1)$
So Total unordered pair is $ = 24$
My Question is , is there is any other Method to calculate the ordered pair in less complex way
because above is very Lengthy method
Help Required
Thanks.
algebra-precalculus number-theory
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
$endgroup$
[1] Number of ordered pair of unequal positive integer solution of $x+y+z = 10$
[2] Number of ordered pair of unequal positive integer solution of $x+y+z+w = 20$
$bf{My; Try}::$ For $(1)$ one:: Here $x,y,z>0$ and $x,y,zin mathbb{Z^{+}}$
$bullet $ If $x=1$, Then $y+z=9$, So $(y,z) = (2,7);,(3,6);,(4,5);(5,4);,(6,3);,(7,2)$
$bullet $ If $x=2$ Then $y+z=8$, So $(y,z) = (1,7);,(3,5);,(5,3);(7,1)$
$bullet $ If $x=3$ Then $y+z=7$, So $(y,z) = (1,6);,(2,5);,(5,2);(6,1)$
$bullet $ If $x=4$ Then $y+z=6$, So $(y,z) = (1,5);,(5,1)$.
$bullet $ If $x=5$ Then $y+z=5$, So $(y,z) = (1,4);,(2,3);,(3,2);(4,1)$
$bullet $ If $x=6$ Then $y+z=4$, So $(y,z) = (1,3);,(3,1)$
$bullet $ If $x=7$ Then $y+z=3$, So $(y,z) = (1,2);,(2,1)$
So Total unordered pair is $ = 24$
My Question is , is there is any other Method to calculate the ordered pair in less complex way
because above is very Lengthy method
Help Required
Thanks.
algebra-precalculus number-theory
algebra-precalculus number-theory
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
edited Jan 26 '14 at 1:42
Matthew Conroy
10.3k32836
10.3k32836
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
asked Jan 25 '14 at 18:09
juantheronjuantheron
34.3k1147142
34.3k1147142
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) le S
$$
so $3x+3 le S$, i.e., $xle (S/3)-1$.
As well, $y+(y+1)le S-x$, i.e., $2y+1 le S-x$, i.e. $y le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=sum_{x:1 le x le frac{S}{3}-1} sum_{y:x+1le y le frac{S-x-1}{2}} 1
= sum_{x:1 le x le frac{S}{3}-1} left( leftlfloor frac{S-x-1}{2}rightrfloor -xright)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=mathrm{round}left( frac{(S-3)^2}{12} right)$$
and I've verified that for $6le Sle 10000$, so it's pretty surely correct. A proof is probably among the many references at that OEIS entry.
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) le S
$$
so $3x+3 le S$, i.e., $xle (S/3)-1$.
As well, $y+(y+1)le S-x$, i.e., $2y+1 le S-x$, i.e. $y le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=sum_{x:1 le x le frac{S}{3}-1} sum_{y:x+1le y le frac{S-x-1}{2}} 1
= sum_{x:1 le x le frac{S}{3}-1} left( leftlfloor frac{S-x-1}{2}rightrfloor -xright)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=mathrm{round}left( frac{(S-3)^2}{12} right)$$
and I've verified that for $6le Sle 10000$, so it's pretty surely correct. A proof is probably among the many references at that OEIS entry.
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
$endgroup$
add a comment |
$begingroup$
For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) le S
$$
so $3x+3 le S$, i.e., $xle (S/3)-1$.
As well, $y+(y+1)le S-x$, i.e., $2y+1 le S-x$, i.e. $y le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=sum_{x:1 le x le frac{S}{3}-1} sum_{y:x+1le y le frac{S-x-1}{2}} 1
= sum_{x:1 le x le frac{S}{3}-1} left( leftlfloor frac{S-x-1}{2}rightrfloor -xright)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=mathrm{round}left( frac{(S-3)^2}{12} right)$$
and I've verified that for $6le Sle 10000$, so it's pretty surely correct. A proof is probably among the many references at that OEIS entry.
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
$endgroup$
add a comment |
$begingroup$
For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) le S
$$
so $3x+3 le S$, i.e., $xle (S/3)-1$.
As well, $y+(y+1)le S-x$, i.e., $2y+1 le S-x$, i.e. $y le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=sum_{x:1 le x le frac{S}{3}-1} sum_{y:x+1le y le frac{S-x-1}{2}} 1
= sum_{x:1 le x le frac{S}{3}-1} left( leftlfloor frac{S-x-1}{2}rightrfloor -xright)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=mathrm{round}left( frac{(S-3)^2}{12} right)$$
and I've verified that for $6le Sle 10000$, so it's pretty surely correct. A proof is probably among the many references at that OEIS entry.
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
$endgroup$
For the first question, let's look at this for a general sum, $S$. We seek triples of unequal positive integers $(x,y,z)$ with $x+y+z=S$. We can count them by assuming $x<y<z$, and multiply the resulting count by $3!=6$.
Since we are assuming $x<y<z$, we must have
$$
x+(x+1)+(x+2) le S
$$
so $3x+3 le S$, i.e., $xle (S/3)-1$.
As well, $y+(y+1)le S-x$, i.e., $2y+1 le S-x$, i.e. $y le (S-x-1)/2.$
Once $x$ and $y$ are chosen, this forces the value of $z$, and so the quantity you seek is $6$ times
$$
f(S)=sum_{x:1 le x le frac{S}{3}-1} sum_{y:x+1le y le frac{S-x-1}{2}} 1
= sum_{x:1 le x le frac{S}{3}-1} left( leftlfloor frac{S-x-1}{2}rightrfloor -xright)
$$
Calculating this for various $S$, starting with $S=6$, we have the sequence
$$1,1,2,3,4,5,7,8,10,12,14,16,19,21,24,27,30,33,37,40,44,... .$$
For your $S=10$, we get the value $4$, which when multiplied by $6$ yields your 24.
This sequence appears to be (a shift of) http://oeis.org/A001399, and from the entry there, it seems that $$f(S)=mathrm{round}left( frac{(S-3)^2}{12} right)$$
and I've verified that for $6le Sle 10000$, so it's pretty surely correct. A proof is probably among the many references at that OEIS entry.
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
edited Dec 7 '18 at 21:47
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
answered Jan 26 '14 at 1:38
Matthew ConroyMatthew Conroy
10.3k32836
10.3k32836
add a comment |
add a comment |
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