Alchemists series; What is it?

[grayson]

Okay, Here is a number theory concept i thought about

Lets say you have two imaginary colors, and you have to calculate the number of mixtures between those (The reason it is imaginary is because to many mixtures makes it brown)

You have two colors. So how many mixtures do they have possibly? Well one!

So what about three? Well three!

But when you get to four, that is when it gets hard.

So if it keeps going on forever, what numbers do you get?

[exchemist]

9 hours ago, grayson said:

Okay, Here is a number theory concept i thought about

Lets say you have two imaginary colors, and you have to calculate the number of mixtures between those (The reason it is imaginary is because to many mixtures makes it brown)

You have two colors. So how many mixtures do they have possibly? Well one!

So what about three? Well three!

But when you get to four, that is when it gets hard.

So if it keeps going on forever, what numbers do you get?

Look up mathematical combinations. 

[John Cuthber]

10 hours ago, grayson said:

So what about three? Well three!

A+B
A+C
B+C

A+B+C
makes 4
If you consider "A + nothing" to be a mixture
You have another 3 "mixtures".

And You can sort of include "nothing + nothing"
Which give you 8 altogether.

Imagine you introduce D into the system
You can either add it, or not add it, to all the previous mixtures.
So that doubles the number of possibilities.
So for 4 materials there are 16 possible mixtures- and in general you get 2^n mixtures of n substances.

But the alchemists (and chemists) distinguish between different compounds, even when they contain the same elements, eg FeO , Fe₂O₃ and Fe₃O₄


 

[Genady]

11 hours ago, grayson said:

You have two colors. So how many mixtures do they have possibly? Well one!

No. Infinite. Depends on the proportion of the colors in the mixture.

But if you are asking, how many different pairs can one make out of n different items, then the answer is n(n-1)/2.

Edited August 22, 2023 by Genady

[grayson]

On 8/22/2023 at 4:45 AM, Genady said:

No. Infinite. Depends on the proportion of the colors in the mixture.

But if you are asking, how many different pairs can one make out of n different items, then the answer is n(n-1)/2.

That is good. But an easier way to write it down on a calculator is nCr(4, 2). than you would have to do everything else (Sorry, bad explaining. You have to find the triple mixtures and so on). so it would be:

nCr(4, 1)+nCr(4, 2)+nCr(4, 3)+nCr(4, 4) = 15

I mean binomial coefficients

[Genady]

6 minutes ago, grayson said:

That is good. But an easier way to write it down on a calculator is nCr(4, 2). than you would have to do everything else (Sorry, bad explaining. You have to find the triple mixtures and so on). so it would be:

nCr(4, 1)+nCr(4, 2)+nCr(4, 3)+nCr(4, 4) = 15

Well, there is one more possibility, nCr(4,0) = 1.

If you count all the 'mixtures', then, instead of summation as you did above, you better calculate a simple total directly, which is 2ⁿ as @John Cuthber has explained earlier. And if you don't want to count the 'empty mixture' nCr(4,0), then the answer is simply 2ⁿ-1.

[grayson]

19 minutes ago, Genady said:

Well, there is one more possibility, nCr(4,0) = 1.

If you count all the 'mixtures', then, instead of summation as you did above, you better calculate a simple total directly, which is 2ⁿ as @John Cuthber has explained earlier. And if you don't want to count the 'empty mixture' nCr(4,0), then the answer is simply 2ⁿ-1.

Oh, that is good! I was hoping I could find a way to shorten it. And no, I don't want to count the zero. Just because these are imaginary number doesn't mean they aren;t physical

Uh, oh. I put in three and it came out seven. It should be 4

Nevermind I'm just stupid

 

[Genady]

14 minutes ago, grayson said:

I put in three and it came out seven. It should be 4

No, it should be 7, not 4.

[grayson]

I know. I just realized I have to not count the original colors. Just the mixtures. So it would really be 2ⁿ-(n+1)