What is the nmber of different positions of N elements in D positions
$begingroup$
I am looking for a general formula that would give me the amount of different P positions that N elements can take with D positions. This N elements are identical so, for exemple (a,b) is the same as (b,a).
Exemple
For N = 2 and D = 4
we have :
[1,1,0,0]
[1,0,1,0]
[1,0,0,1]
[0,1,1,0]
[0,1,0,1]
[0,0,1,1]
wich gives us P = 6 possibilities !
What I tried
N= 1 is rather easy, and I found P = D
for N = 2 I found something like P = $$sum_{i=1}^{D-1} i$$
But when I get to N = 3, it gets tricky and I can't really figure it out. There seems to be something around a sum of a sum, or sum mutliplied by a sum, but I'm not really sure...
There seems to be a pattern, wich make me think that there is a general formula to answer this question. And my math background is not good enough to be able to ask the question in a more former way.
My first goal was to find the solution for N = 4 for and D = 24 for a completely unrelated subject (I'm doing a research project in Neuroscience, wich can explain my poor background in math). But I got stuck at N= 3 and I finelly got intrigued by the question and tried to find a more general answer.
Supplementary Data
I took the time to count every possibilities, in an attempt to understand the pattern. If you are interested, this is the series for an incremental amount of D starting at 1
N=1 : [1,2,3,4,5,6...]
N=2 : [0,1,3,6,10,15...]
N=3 : [0,0,1,4,10,20...]
N=4 : [0,0,0,1,5...]
Thanks in advance for the help !
combinatorics
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
asked Jan 8 at 19:26
HornycarHornycar
182
$endgroup$
add a comment |
$begingroup$
I am looking for a general formula that would give me the amount of different P positions that N elements can take with D positions. This N elements are identical so, for exemple (a,b) is the same as (b,a).
Exemple
For N = 2 and D = 4
we have :
[1,1,0,0]
[1,0,1,0]
[1,0,0,1]
[0,1,1,0]
[0,1,0,1]
[0,0,1,1]
wich gives us P = 6 possibilities !
What I tried
N= 1 is rather easy, and I found P = D
for N = 2 I found something like P = $$sum_{i=1}^{D-1} i$$
But when I get to N = 3, it gets tricky and I can't really figure it out. There seems to be something around a sum of a sum, or sum mutliplied by a sum, but I'm not really sure...
There seems to be a pattern, wich make me think that there is a general formula to answer this question. And my math background is not good enough to be able to ask the question in a more former way.
My first goal was to find the solution for N = 4 for and D = 24 for a completely unrelated subject (I'm doing a research project in Neuroscience, wich can explain my poor background in math). But I got stuck at N= 3 and I finelly got intrigued by the question and tried to find a more general answer.
Supplementary Data
I took the time to count every possibilities, in an attempt to understand the pattern. If you are interested, this is the series for an incremental amount of D starting at 1
N=1 : [1,2,3,4,5,6...]
N=2 : [0,1,3,6,10,15...]
N=3 : [0,0,1,4,10,20...]
N=4 : [0,0,0,1,5...]
Thanks in advance for the help !
combinatorics
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
asked Jan 8 at 19:26
HornycarHornycar
182
$endgroup$
3
$begingroup$
I do not fully understand the types of objects you are trying to count as you didn't explain it very well, however the numbers you list at the end are simply Binomial coefficients. $binom{3}{3}=1,binom{4}{3}=4,binom{5}{3}=10,binom{6}{3}=20,binom{7}{3}=35,dots$. It looks like you are simply looking for $binom{D}{N}$.
$endgroup$
– JMoravitz
Jan 8 at 19:32
$begingroup$
I guess it wasn't clearly explained because I don't know how to explain it in a formal way. It can be boxes, as long as they are identical and you can't make the difference between them. If the Binomial coefficients correspond to the lists, I suppose that this is the answer ! I'm going to look a little bit more to the wikipedia page to make sure but it looks like it !
$endgroup$
– Hornycar
Jan 8 at 19:43
add a comment |
$begingroup$
I am looking for a general formula that would give me the amount of different P positions that N elements can take with D positions. This N elements are identical so, for exemple (a,b) is the same as (b,a).
Exemple
For N = 2 and D = 4
we have :
[1,1,0,0]
[1,0,1,0]
[1,0,0,1]
[0,1,1,0]
[0,1,0,1]
[0,0,1,1]
wich gives us P = 6 possibilities !
What I tried
N= 1 is rather easy, and I found P = D
for N = 2 I found something like P = $$sum_{i=1}^{D-1} i$$
But when I get to N = 3, it gets tricky and I can't really figure it out. There seems to be something around a sum of a sum, or sum mutliplied by a sum, but I'm not really sure...
There seems to be a pattern, wich make me think that there is a general formula to answer this question. And my math background is not good enough to be able to ask the question in a more former way.
My first goal was to find the solution for N = 4 for and D = 24 for a completely unrelated subject (I'm doing a research project in Neuroscience, wich can explain my poor background in math). But I got stuck at N= 3 and I finelly got intrigued by the question and tried to find a more general answer.
Supplementary Data
I took the time to count every possibilities, in an attempt to understand the pattern. If you are interested, this is the series for an incremental amount of D starting at 1
N=1 : [1,2,3,4,5,6...]
N=2 : [0,1,3,6,10,15...]
N=3 : [0,0,1,4,10,20...]
N=4 : [0,0,0,1,5...]
Thanks in advance for the help !
combinatorics
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
asked Jan 8 at 19:26
HornycarHornycar
182
$endgroup$
I am looking for a general formula that would give me the amount of different P positions that N elements can take with D positions. This N elements are identical so, for exemple (a,b) is the same as (b,a).
Exemple
For N = 2 and D = 4
we have :
[1,1,0,0]
[1,0,1,0]
[1,0,0,1]
[0,1,1,0]
[0,1,0,1]
[0,0,1,1]
wich gives us P = 6 possibilities !
What I tried
N= 1 is rather easy, and I found P = D
for N = 2 I found something like P = $$sum_{i=1}^{D-1} i$$
But when I get to N = 3, it gets tricky and I can't really figure it out. There seems to be something around a sum of a sum, or sum mutliplied by a sum, but I'm not really sure...
There seems to be a pattern, wich make me think that there is a general formula to answer this question. And my math background is not good enough to be able to ask the question in a more former way.
My first goal was to find the solution for N = 4 for and D = 24 for a completely unrelated subject (I'm doing a research project in Neuroscience, wich can explain my poor background in math). But I got stuck at N= 3 and I finelly got intrigued by the question and tried to find a more general answer.
Supplementary Data
I took the time to count every possibilities, in an attempt to understand the pattern. If you are interested, this is the series for an incremental amount of D starting at 1
N=1 : [1,2,3,4,5,6...]
N=2 : [0,1,3,6,10,15...]
N=3 : [0,0,1,4,10,20...]
N=4 : [0,0,0,1,5...]
Thanks in advance for the help !
combinatorics
combinatorics
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
asked Jan 8 at 19:26
HornycarHornycar
182
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
asked Jan 8 at 19:26
HornycarHornycar
182
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
edited Jan 9 at 23:35
Ted Shifrin
63.1k44489
63.1k44489
asked Jan 8 at 19:26
HornycarHornycar
182
asked Jan 8 at 19:26
HornycarHornycar
182
asked Jan 8 at 19:26
HornycarHornycar
182
182
3
$begingroup$
I do not fully understand the types of objects you are trying to count as you didn't explain it very well, however the numbers you list at the end are simply Binomial coefficients. $binom{3}{3}=1,binom{4}{3}=4,binom{5}{3}=10,binom{6}{3}=20,binom{7}{3}=35,dots$. It looks like you are simply looking for $binom{D}{N}$.
$endgroup$
– JMoravitz
Jan 8 at 19:32
$begingroup$
I guess it wasn't clearly explained because I don't know how to explain it in a formal way. It can be boxes, as long as they are identical and you can't make the difference between them. If the Binomial coefficients correspond to the lists, I suppose that this is the answer ! I'm going to look a little bit more to the wikipedia page to make sure but it looks like it !
$endgroup$
– Hornycar
Jan 8 at 19:43
add a comment |
3
$begingroup$
I do not fully understand the types of objects you are trying to count as you didn't explain it very well, however the numbers you list at the end are simply Binomial coefficients. $binom{3}{3}=1,binom{4}{3}=4,binom{5}{3}=10,binom{6}{3}=20,binom{7}{3}=35,dots$. It looks like you are simply looking for $binom{D}{N}$.
$endgroup$
– JMoravitz
Jan 8 at 19:32
$begingroup$
I guess it wasn't clearly explained because I don't know how to explain it in a formal way. It can be boxes, as long as they are identical and you can't make the difference between them. If the Binomial coefficients correspond to the lists, I suppose that this is the answer ! I'm going to look a little bit more to the wikipedia page to make sure but it looks like it !
$endgroup$
– Hornycar
Jan 8 at 19:43
3
3
$begingroup$
I do not fully understand the types of objects you are trying to count as you didn't explain it very well, however the numbers you list at the end are simply Binomial coefficients. $binom{3}{3}=1,binom{4}{3}=4,binom{5}{3}=10,binom{6}{3}=20,binom{7}{3}=35,dots$. It looks like you are simply looking for $binom{D}{N}$.
$endgroup$
– JMoravitz
Jan 8 at 19:32
$begingroup$
I do not fully understand the types of objects you are trying to count as you didn't explain it very well, however the numbers you list at the end are simply Binomial coefficients. $binom{3}{3}=1,binom{4}{3}=4,binom{5}{3}=10,binom{6}{3}=20,binom{7}{3}=35,dots$. It looks like you are simply looking for $binom{D}{N}$.
$endgroup$
– JMoravitz
Jan 8 at 19:32
$begingroup$
I guess it wasn't clearly explained because I don't know how to explain it in a formal way. It can be boxes, as long as they are identical and you can't make the difference between them. If the Binomial coefficients correspond to the lists, I suppose that this is the answer ! I'm going to look a little bit more to the wikipedia page to make sure but it looks like it !
$endgroup$
– Hornycar
Jan 8 at 19:43
$begingroup$
I guess it wasn't clearly explained because I don't know how to explain it in a formal way. It can be boxes, as long as they are identical and you can't make the difference between them. If the Binomial coefficients correspond to the lists, I suppose that this is the answer ! I'm going to look a little bit more to the wikipedia page to make sure but it looks like it !
$endgroup$
– Hornycar
Jan 8 at 19:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you don't distinguish the elements, you'll use combinations.
Taking your example:
$N$ identical (indistinguishable) objects have to be placed in $N$ boxes chosen from the total of $D$ boxes. What you need is to chose $N$ boxes. The number of possibilities is expressed as the number of $N$-combinations and equals
$$binom{D}{N}=frac{D!}{(D-N)!;N!}$$
In particular, if $;N=4,;D=24;$ there are
$$binom{24}{4}=23times22times21=10626$$ possibilities.
Note:
The numbers you give for different values of $N$ in strings are exactly so, e.g. $binom{5}{3}=10.$
answered Jan 8 at 21:41
user376343user376343
3,3532825
$endgroup$
add a comment |
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$begingroup$
If you don't distinguish the elements, you'll use combinations.
Taking your example:
$N$ identical (indistinguishable) objects have to be placed in $N$ boxes chosen from the total of $D$ boxes. What you need is to chose $N$ boxes. The number of possibilities is expressed as the number of $N$-combinations and equals
$$binom{D}{N}=frac{D!}{(D-N)!;N!}$$
In particular, if $;N=4,;D=24;$ there are
$$binom{24}{4}=23times22times21=10626$$ possibilities.
Note:
The numbers you give for different values of $N$ in strings are exactly so, e.g. $binom{5}{3}=10.$
answered Jan 8 at 21:41
user376343user376343
3,3532825
$endgroup$
add a comment |
$begingroup$
If you don't distinguish the elements, you'll use combinations.
Taking your example:
$N$ identical (indistinguishable) objects have to be placed in $N$ boxes chosen from the total of $D$ boxes. What you need is to chose $N$ boxes. The number of possibilities is expressed as the number of $N$-combinations and equals
$$binom{D}{N}=frac{D!}{(D-N)!;N!}$$
In particular, if $;N=4,;D=24;$ there are
$$binom{24}{4}=23times22times21=10626$$ possibilities.
Note:
The numbers you give for different values of $N$ in strings are exactly so, e.g. $binom{5}{3}=10.$
answered Jan 8 at 21:41
user376343user376343
3,3532825
$endgroup$
add a comment |
$begingroup$
If you don't distinguish the elements, you'll use combinations.
Taking your example:
$N$ identical (indistinguishable) objects have to be placed in $N$ boxes chosen from the total of $D$ boxes. What you need is to chose $N$ boxes. The number of possibilities is expressed as the number of $N$-combinations and equals
$$binom{D}{N}=frac{D!}{(D-N)!;N!}$$
In particular, if $;N=4,;D=24;$ there are
$$binom{24}{4}=23times22times21=10626$$ possibilities.
Note:
The numbers you give for different values of $N$ in strings are exactly so, e.g. $binom{5}{3}=10.$
answered Jan 8 at 21:41
user376343user376343
3,3532825
$endgroup$
If you don't distinguish the elements, you'll use combinations.
Taking your example:
$N$ identical (indistinguishable) objects have to be placed in $N$ boxes chosen from the total of $D$ boxes. What you need is to chose $N$ boxes. The number of possibilities is expressed as the number of $N$-combinations and equals
$$binom{D}{N}=frac{D!}{(D-N)!;N!}$$
In particular, if $;N=4,;D=24;$ there are
$$binom{24}{4}=23times22times21=10626$$ possibilities.
Note:
The numbers you give for different values of $N$ in strings are exactly so, e.g. $binom{5}{3}=10.$
answered Jan 8 at 21:41
user376343user376343
3,3532825
edited Jan 9 at 17:55
answered Jan 8 at 21:41
user376343user376343
3,3532825
answered Jan 8 at 21:41
user376343user376343
3,3532825
answered Jan 8 at 21:41
user376343user376343
3,3532825
3,3532825
add a comment |
add a comment |
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$begingroup$
I do not fully understand the types of objects you are trying to count as you didn't explain it very well, however the numbers you list at the end are simply Binomial coefficients. $binom{3}{3}=1,binom{4}{3}=4,binom{5}{3}=10,binom{6}{3}=20,binom{7}{3}=35,dots$. It looks like you are simply looking for $binom{D}{N}$.
$endgroup$
– JMoravitz
Jan 8 at 19:32
$begingroup$
I guess it wasn't clearly explained because I don't know how to explain it in a formal way. It can be boxes, as long as they are identical and you can't make the difference between them. If the Binomial coefficients correspond to the lists, I suppose that this is the answer ! I'm going to look a little bit more to the wikipedia page to make sure but it looks like it !
$endgroup$
– Hornycar
Jan 8 at 19:43