possible combinations

[yarnedia]

I need to find possible combinations for the following example:

a*x + b*y + c*z = t
p = 10

Where
a = 0
b = 5
c = 10
t = 30

x + y + z has to = p

a, b, c, t, and p would be variables that I would specify. The numbers above are just examples. What I would need to see as output would be:

7a, 0b, 3c
4a, 6b, 0c
5a, 4b, 1c
etc...

[yarnedia]

any help is greatly appreciated!

[shg]

You have 7 unknowns and 5 equations. There are an infinite number of solutions.

[yarnedia]

No, there are only 3 unkowns. Values for a,b,c,t,and p are specified. For example using this particular example the potential solutions are:

6y and 4x
4y, 1z and 5x
2y, 2z and 6x
3z and 7x

All of these combinations result in t=30 and x+y+z=10 (or p)

[shg]

OK, there are three unknowns -- still an infinite number of solutions. Are x, y and z constrained to be non-negative?

[yarnedia]

Yes, xyz cannot be negative.

To be clear, I am not looking for the number of possibilities. I am looking for actual possible combinations when the variables of a,b,c,t,p are assigned values (> 0)

[shg]

So,

x + y + z = p => z = p - x - y

For x = 0, there are p+1 possibilities for y and z
For x = 1, there are p possibilities for y and z
For x = 2, there are p-1 possibilities for y and z
For x = 3, there are p-2 possibilities for y and z
...
For x = p, there is 1 possibility for y and z

The sum of {i, i+1, i+2, ..., P+1} is (P+1)*(P+2)/2

For P=0, 1, 2, 3, 4, ..., combinations = 1, 3, 6, 10, 15, ... (which are the triangular numbers starting at 1)

To list them,

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[yarnedia]

excellent, this did the trick. Thanks!!!

[shg]

You’re welcome. Would you please mark the thread as Solved?

Click the Edit button on your first post in the thread

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