Mixing
Calculating no. of solutions within some bounds
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Find the total number of non-negative integral ordered triplets for-
$$x_1+x_2+x_3=11$$
under the bounds,
$x_1in (2,6)$ and
$x_2 in (3,7)$.
I was generally able to solve such problems involving bounds by introducing a new variable but that was only in case of a single boundary condition.
How do I proceed when there are two?
combinatorics discrete-mathematics
edited yesterday
Tianlalu
2,574632
asked yesterday
Sameer Thakur
142
add a comment |
up vote
0
down vote
favorite
Find the total number of non-negative integral ordered triplets for-
$$x_1+x_2+x_3=11$$
under the bounds,
$x_1in (2,6)$ and
$x_2 in (3,7)$.
I was generally able to solve such problems involving bounds by introducing a new variable but that was only in case of a single boundary condition.
How do I proceed when there are two?
combinatorics discrete-mathematics
edited yesterday
Tianlalu
2,574632
asked yesterday
Sameer Thakur
142
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the total number of non-negative integral ordered triplets for-
$$x_1+x_2+x_3=11$$
under the bounds,
$x_1in (2,6)$ and
$x_2 in (3,7)$.
I was generally able to solve such problems involving bounds by introducing a new variable but that was only in case of a single boundary condition.
How do I proceed when there are two?
combinatorics discrete-mathematics
edited yesterday
Tianlalu
2,574632
asked yesterday
Sameer Thakur
142
Find the total number of non-negative integral ordered triplets for-
$$x_1+x_2+x_3=11$$
under the bounds,
$x_1in (2,6)$ and
$x_2 in (3,7)$.
I was generally able to solve such problems involving bounds by introducing a new variable but that was only in case of a single boundary condition.
How do I proceed when there are two?
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited yesterday
Tianlalu
2,574632
asked yesterday
Sameer Thakur
142
edited yesterday
Tianlalu
2,574632
asked yesterday
Sameer Thakur
142
edited yesterday
Tianlalu
2,574632
edited yesterday
Tianlalu
2,574632
edited yesterday
Tianlalu
2,574632
2,574632
asked yesterday
Sameer Thakur
142
asked yesterday
Sameer Thakur
142
asked yesterday
Sameer Thakur
142
142
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
I have figured out an approach using Multinomial Theorem-
We can find the coefficient of x^11 in (x^3+x^4+x^5)(x^4+x^5++x^6)(x^1+x^2+x^3+x^4).
This comes out to be 8.
answered yesterday
Sameer Thakur
142
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
add a comment |
up vote
0
down vote
This can be solved using Bose-Einstein, which is also known as Bars and Stars https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics).
In your case, you have 11 dots, and basically you want to put this 11 dots into 3 boxes. You only care how many dots are going in each box. For this you can use:
$$binom{n + k - 1}{k-1}$$ and in your case: $$binom{11 + 3 - 1}{2}$$ This is including the solutions where you have 0 as possible value for some variables, so you might want to exclude those.
First let's get all the possible solutions that match the lower bound.
Smallest value for x1 is 3 and for x2 is 4. We can mark x1 and x2 as:
$$x1 = y1 + 3\
x2 = y2 + 4
$$
The new equation becomes: $$ y1 + 3 + y2 + 4 + x3 = 11 <=> y1 + y2 + x3 = 4 $$
Now all the possible values are:
$$ binom{4 + 3 - 1}{3 - 1}$$
From this ones you need to substract all the values that have either x1 > 6 or x2 > 7. (They can't be both greater since the sum would be over the boundaries). So you can replace x1 with y1 + 6 and x2 with y2 + 7 and apply the same principle. After that you can substract these results from the total.
answered yesterday
Erik Cristian Seulean
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have edited the question.
– Sameer Thakur
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I have figured out an approach using Multinomial Theorem-
We can find the coefficient of x^11 in (x^3+x^4+x^5)(x^4+x^5++x^6)(x^1+x^2+x^3+x^4).
This comes out to be 8.
answered yesterday
Sameer Thakur
142
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
add a comment |
up vote
0
down vote
I have figured out an approach using Multinomial Theorem-
We can find the coefficient of x^11 in (x^3+x^4+x^5)(x^4+x^5++x^6)(x^1+x^2+x^3+x^4).
This comes out to be 8.
answered yesterday
Sameer Thakur
142
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
I have figured out an approach using Multinomial Theorem-
We can find the coefficient of x^11 in (x^3+x^4+x^5)(x^4+x^5++x^6)(x^1+x^2+x^3+x^4).
This comes out to be 8.
answered yesterday
Sameer Thakur
142
I have figured out an approach using Multinomial Theorem-
We can find the coefficient of x^11 in (x^3+x^4+x^5)(x^4+x^5++x^6)(x^1+x^2+x^3+x^4).
This comes out to be 8.
answered yesterday
Sameer Thakur
142
answered yesterday
Sameer Thakur
142
answered yesterday
Sameer Thakur
142
answered yesterday
Sameer Thakur
142
142
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
add a comment |
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
Can someone guide me as to whether this is a correct approach?
– Sameer Thakur
yesterday
add a comment |
up vote
0
down vote
This can be solved using Bose-Einstein, which is also known as Bars and Stars https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics).
In your case, you have 11 dots, and basically you want to put this 11 dots into 3 boxes. You only care how many dots are going in each box. For this you can use:
$$binom{n + k - 1}{k-1}$$ and in your case: $$binom{11 + 3 - 1}{2}$$ This is including the solutions where you have 0 as possible value for some variables, so you might want to exclude those.
First let's get all the possible solutions that match the lower bound.
Smallest value for x1 is 3 and for x2 is 4. We can mark x1 and x2 as:
$$x1 = y1 + 3\
x2 = y2 + 4
$$
The new equation becomes: $$ y1 + 3 + y2 + 4 + x3 = 11 <=> y1 + y2 + x3 = 4 $$
Now all the possible values are:
$$ binom{4 + 3 - 1}{3 - 1}$$
From this ones you need to substract all the values that have either x1 > 6 or x2 > 7. (They can't be both greater since the sum would be over the boundaries). So you can replace x1 with y1 + 6 and x2 with y2 + 7 and apply the same principle. After that you can substract these results from the total.
answered yesterday
Erik Cristian Seulean
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have edited the question.
– Sameer Thakur
yesterday
add a comment |
up vote
0
down vote
This can be solved using Bose-Einstein, which is also known as Bars and Stars https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics).
In your case, you have 11 dots, and basically you want to put this 11 dots into 3 boxes. You only care how many dots are going in each box. For this you can use:
$$binom{n + k - 1}{k-1}$$ and in your case: $$binom{11 + 3 - 1}{2}$$ This is including the solutions where you have 0 as possible value for some variables, so you might want to exclude those.
First let's get all the possible solutions that match the lower bound.
Smallest value for x1 is 3 and for x2 is 4. We can mark x1 and x2 as:
$$x1 = y1 + 3\
x2 = y2 + 4
$$
The new equation becomes: $$ y1 + 3 + y2 + 4 + x3 = 11 <=> y1 + y2 + x3 = 4 $$
Now all the possible values are:
$$ binom{4 + 3 - 1}{3 - 1}$$
From this ones you need to substract all the values that have either x1 > 6 or x2 > 7. (They can't be both greater since the sum would be over the boundaries). So you can replace x1 with y1 + 6 and x2 with y2 + 7 and apply the same principle. After that you can substract these results from the total.
answered yesterday
Erik Cristian Seulean
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have edited the question.
– Sameer Thakur
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
This can be solved using Bose-Einstein, which is also known as Bars and Stars https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics).
In your case, you have 11 dots, and basically you want to put this 11 dots into 3 boxes. You only care how many dots are going in each box. For this you can use:
$$binom{n + k - 1}{k-1}$$ and in your case: $$binom{11 + 3 - 1}{2}$$ This is including the solutions where you have 0 as possible value for some variables, so you might want to exclude those.
First let's get all the possible solutions that match the lower bound.
Smallest value for x1 is 3 and for x2 is 4. We can mark x1 and x2 as:
$$x1 = y1 + 3\
x2 = y2 + 4
$$
The new equation becomes: $$ y1 + 3 + y2 + 4 + x3 = 11 <=> y1 + y2 + x3 = 4 $$
Now all the possible values are:
$$ binom{4 + 3 - 1}{3 - 1}$$
From this ones you need to substract all the values that have either x1 > 6 or x2 > 7. (They can't be both greater since the sum would be over the boundaries). So you can replace x1 with y1 + 6 and x2 with y2 + 7 and apply the same principle. After that you can substract these results from the total.
answered yesterday
Erik Cristian Seulean
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This can be solved using Bose-Einstein, which is also known as Bars and Stars https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics).
In your case, you have 11 dots, and basically you want to put this 11 dots into 3 boxes. You only care how many dots are going in each box. For this you can use:
$$binom{n + k - 1}{k-1}$$ and in your case: $$binom{11 + 3 - 1}{2}$$ This is including the solutions where you have 0 as possible value for some variables, so you might want to exclude those.
First let's get all the possible solutions that match the lower bound.
Smallest value for x1 is 3 and for x2 is 4. We can mark x1 and x2 as:
$$x1 = y1 + 3\
x2 = y2 + 4
$$
The new equation becomes: $$ y1 + 3 + y2 + 4 + x3 = 11 <=> y1 + y2 + x3 = 4 $$
Now all the possible values are:
$$ binom{4 + 3 - 1}{3 - 1}$$
From this ones you need to substract all the values that have either x1 > 6 or x2 > 7. (They can't be both greater since the sum would be over the boundaries). So you can replace x1 with y1 + 6 and x2 with y2 + 7 and apply the same principle. After that you can substract these results from the total.
answered yesterday
Erik Cristian Seulean
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
Erik Cristian Seulean
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Erik Cristian Seulean
385
answered yesterday
Erik Cristian Seulean
385
385
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Erik Cristian Seulean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have edited the question.
– Sameer Thakur
yesterday
add a comment |
I have edited the question.
– Sameer Thakur
yesterday
I have edited the question.
– Sameer Thakur
yesterday
I have edited the question.
– Sameer Thakur
yesterday
add a comment |
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