Number ways of arranging dots and lines with restriction
$begingroup$
I have this problem where I need to calculate the following:
For x dots and y lines, how many arrangements are possible such that no more than n dots are together?
Can anyone help me out?
combinatorics discrete-mathematics permutations
asked Dec 13 '18 at 4:03
HarryHarry
12
$endgroup$
add a comment |
$begingroup$
I have this problem where I need to calculate the following:
For x dots and y lines, how many arrangements are possible such that no more than n dots are together?
Can anyone help me out?
combinatorics discrete-mathematics permutations
asked Dec 13 '18 at 4:03
HarryHarry
12
$endgroup$
add a comment |
$begingroup$
I have this problem where I need to calculate the following:
For x dots and y lines, how many arrangements are possible such that no more than n dots are together?
Can anyone help me out?
combinatorics discrete-mathematics permutations
asked Dec 13 '18 at 4:03
HarryHarry
12
$endgroup$
I have this problem where I need to calculate the following:
For x dots and y lines, how many arrangements are possible such that no more than n dots are together?
Can anyone help me out?
combinatorics discrete-mathematics permutations
combinatorics discrete-mathematics permutations
asked Dec 13 '18 at 4:03
HarryHarry
12
asked Dec 13 '18 at 4:03
HarryHarry
12
edited Dec 13 '18 at 4:46
Harry
asked Dec 13 '18 at 4:03
HarryHarry
12
asked Dec 13 '18 at 4:03
HarryHarry
12
asked Dec 13 '18 at 4:03
HarryHarry
12
12
add a comment |
add a comment |
1 Answer
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$begingroup$
First put all the lines, like this | | | | |
Then, out of y lines,$y+1$ gaps will be created (including end spaces). Maximum number of dots in each gap is $n$.
Now, let the number of dots in $i$th gap is $x_i$.
Required number of ways is the number of non-negative solutions of the equation
$$x_1+x_2+...+x_{y+1}=x$$
Where $x_ileq n forall i$, now you can easily do it by finding the coefficient of $m^x$ in the expansion of $(1+m+m^2+...+m^n)^{y+1}$.
Hope it is helpful:)
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
$endgroup$
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
add a comment |
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$begingroup$
First put all the lines, like this | | | | |
Then, out of y lines,$y+1$ gaps will be created (including end spaces). Maximum number of dots in each gap is $n$.
Now, let the number of dots in $i$th gap is $x_i$.
Required number of ways is the number of non-negative solutions of the equation
$$x_1+x_2+...+x_{y+1}=x$$
Where $x_ileq n forall i$, now you can easily do it by finding the coefficient of $m^x$ in the expansion of $(1+m+m^2+...+m^n)^{y+1}$.
Hope it is helpful:)
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
$endgroup$
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
add a comment |
$begingroup$
First put all the lines, like this | | | | |
Then, out of y lines,$y+1$ gaps will be created (including end spaces). Maximum number of dots in each gap is $n$.
Now, let the number of dots in $i$th gap is $x_i$.
Required number of ways is the number of non-negative solutions of the equation
$$x_1+x_2+...+x_{y+1}=x$$
Where $x_ileq n forall i$, now you can easily do it by finding the coefficient of $m^x$ in the expansion of $(1+m+m^2+...+m^n)^{y+1}$.
Hope it is helpful:)
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
$endgroup$
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
add a comment |
$begingroup$
First put all the lines, like this | | | | |
Then, out of y lines,$y+1$ gaps will be created (including end spaces). Maximum number of dots in each gap is $n$.
Now, let the number of dots in $i$th gap is $x_i$.
Required number of ways is the number of non-negative solutions of the equation
$$x_1+x_2+...+x_{y+1}=x$$
Where $x_ileq n forall i$, now you can easily do it by finding the coefficient of $m^x$ in the expansion of $(1+m+m^2+...+m^n)^{y+1}$.
Hope it is helpful:)
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
$endgroup$
First put all the lines, like this | | | | |
Then, out of y lines,$y+1$ gaps will be created (including end spaces). Maximum number of dots in each gap is $n$.
Now, let the number of dots in $i$th gap is $x_i$.
Required number of ways is the number of non-negative solutions of the equation
$$x_1+x_2+...+x_{y+1}=x$$
Where $x_ileq n forall i$, now you can easily do it by finding the coefficient of $m^x$ in the expansion of $(1+m+m^2+...+m^n)^{y+1}$.
Hope it is helpful:)
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
edited Dec 13 '18 at 5:42
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
answered Dec 13 '18 at 4:13
MartundMartund
1,633213
1,633213
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
add a comment |
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
In the case where x = 4, n = 5 and y = 1, the restriction of n basically doesn't come into effect, so we can just calculate the possibilities the normal way: Arrangements = 6C1 = 6. Using your method, I would get 10C4 = 210, which is not the same. Is there something more which we need to consider with your method?
$endgroup$
– Harry
Dec 13 '18 at 4:34
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
This method treats different “spots” in the same gap as distinct. But they should not be; you’re significantly overcounting.
$endgroup$
– platty
Dec 13 '18 at 4:49
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
$begingroup$
Ok, I have edited it to correct.
$endgroup$
– Martund
Dec 13 '18 at 5:29
add a comment |
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