Let's say you have monomers of glucose(single molecule of glucose). Each glucose molecule has 6 OHs (hydroxyl group) that can react with other OHs on other glucose molecules.

So glucose_1's carbon 1-Oh can react with glucose_2's carbon 1-OH, or carbon 2-OH, ... or carbon6's OH, so on.

Let's say you have a dimer of glucose. The 2 molecules of glucose each have used up 1 OH to form the bond, so they each have 5 available OHs to form the bond a 3rd molecule of glucose.

Think of it as six points on a 6 pointed arrow https://upload.wikimedia.org/wikipedia/commons/thumb/f/f8/19521_Chaos_symbol.svg/1200px-19521_Chaos_symbol.svg.png
Each of the 6 points can connect with another identical 6 pointed arrow. If you add more 6 pointed arrows, you get a scenario where the possible combination of configuration increase exponentially.

For the example of glucose, given 2 molecules of glucose with 6 OHs each, there are 36 possible dimers than can form. (glucose_1 carbon_1-OH can react with 6 OHs on glucose_2. glucose_1 carbon_2-OH can react with 6 OHs on glucose_2 etc for all 6 carbons on glucose_1)

So given the 6 OHs on gluocse_1, you can react it with 6 OHs on glucose_2

But if you consider a trimer of glucose, you might think it's just 6*6*6, but it's actually not, because the 3rd glucose can react with 10 remaining OHs on the glucose dimer using its 6 OHs. It's actually 6*6*10.

And the for the polymer of 4 glucose molecules, the 4th glucose has 15 OHs on the trimer that it can react with using its 6 OHs.

https://www.dummies.com/wp-content/uploads/diabetes-carb-structure.jpg

What is the formal way to think about this problem using combinatorics for arbitrary length n for the number of glucose molecules in a polymer?

I hope I made this problem understandable for you. I lack the math language skills to explain this. Any insights will be helpful.

bo reddude <boreddude003@gmail.com> wrote:
> Let's say you have monomers of glucose(single molecule of glucose).
[snip chem stuff and link to 8-point figure]

> Each of the 6 points can connect with another identical 6 pointed arrow. If you add more 6 pointed arrows, you get a scenario where the possible combination of configuration increase exponentially.
>
> For the example of glucose, given 2 molecules of glucose with 6 OHs each, there are 36 possible dimers than can form. (glucose_1 carbon_1-OH can react with 6 OHs on glucose_2. glucose_1 carbon_2-OH can react with 6 OHs on glucose_2 etc for all 6 carbons on glucose_1)
>
> So given the 6 OHs on gluocse_1, you can react it with 6 OHs on glucose_2
>
> But if you consider a trimer of glucose, you might think it's just 6*6*6, but it's actually not, because the 3rd glucose can react with 10 remaining OHs on the glucose dimer using its 6 OHs. It's actually 6*6*10.
>
> And the for the polymer of 4 glucose molecules, the 4th glucose has 15 OHs on the trimer that it can react with using its 6 OHs.
[snip link]
> What is the formal way to think about this problem using combinatorics for arbitrary length n for the number of glucose molecules in a polymer?

Before I think about 4 molecules, let me ask about the 3-molecule
case, where you say the count is 6*6*10, or 360. My impression is
that you have labeled each molecule (glucose_1, glucose_2, etc but I'm
just going to call them a, b, c) and have labeled the carbons and
their hydroxyls (carbon_1-OH, etc, which I'll refer to as a1...a6,
b1...b6, c1...c6). That is, molecules are distinctly labeled and
C-OHs also distinctly labeled; otherwise, instead of there being 36
a-to-b cases, there would only be 1 case. Also, from your suggested
counts of cases, I'm assuming that there can only be 1 a-to-b
connection at a time. For example if we have a1-b1 connected we
cannot have a2-b2 at the same time. Thus producing branching chains
as illustrated in your second link.

Given such labeling, there are a lot more than 360 3-molecule cases.
Let's consider the case where a has links to b and to c. There are 30
ways to choose which C,OHs of a to connect from. This is 2*(6 2)
which is 2*6*5/2 = 30. [Namely, from a1,a2; a2,a1; a1,a3; a3,a1;
.... a5,a6; a6,a5.] For each of those 30 ways, there are 36 ways to
connect b and c to a; for example, in the a1,a2 case, we have a1-b1,
a2-c1; a1-b1, a2-c2; ... a1-b6, a2-c6. Thus, 30*36 = 1080
arrangements with a-to-b and a-to-c connections.

Now for each of those 1080 arrangements, there are 26 ways to connect
b-to-c. This is 1+5*5, counting 1 no-b-c-connection case and 25 cases
which occupy C-OHs that a didn't connect to. 26*1080 = 28080 cases
with a connected to b and to c, and b-c connected or not.

I haven't worked out the details, but there are a bunch more cases
where a connects to b but not to c, or where it connects to c but not
to b. Eg, 36 a-b cases * 30 b-c cases, and 36 a-c cases * 30 c-b
cases, which if that's all would make the total 28*1080 = 30240.

Anyhow, see if the comments above match up with your chemistry model.
Thanks! - jiw

James Waldby <no@no.no> wrote:
> bo reddude <boreddude003@gmail.com> wrote:
>> Let's say you have monomers of glucose(single molecule of glucose).
> [snip chem stuff and link to 8-point figure]
>
>> Each of the 6 points can connect with another identical 6 pointed arrow. If you add more 6 pointed arrows, you get a scenario where the possible combination of configuration increase exponentially.
>>
>> For the example of glucose, given 2 molecules of glucose with 6 OHs each, there are 36 possible dimers than can form. (glucose_1 carbon_1-OH can react with 6 OHs on glucose_2. glucose_1 carbon_2-OH can react with 6 OHs on glucose_2 etc for all 6 carbons on glucose_1)
>>
>> So given the 6 OHs on gluocse_1, you can react it with 6 OHs on glucose_2
>>
>> But if you consider a trimer of glucose, you might think it's just 6*6*6, but it's actually not, because the 3rd glucose can react with 10 remaining OHs on the glucose dimer using its 6 OHs. It's actually 6*6*10.
>>
>> And the for the polymer of 4 glucose molecules, the 4th glucose has 15 OHs on the trimer that it can react with using its 6 OHs.
> [snip link]
>> What is the formal way to think about this problem using combinatorics for arbitrary length n for the number of glucose molecules in a polymer?
>
> Before I think about 4 molecules, let me ask about the 3-molecule
> case, where you say the count is 6*6*10, or 360. My impression is
> that you have labeled each molecule (glucose_1, glucose_2, etc but I'm
> just going to call them a, b, c) and have labeled the carbons and
> their hydroxyls (carbon_1-OH, etc, which I'll refer to as a1...a6,
> b1...b6, c1...c6). That is, molecules are distinctly labeled and
> C-OHs also distinctly labeled; otherwise, instead of there being 36
> a-to-b cases, there would only be 1 case. Also, from your suggested
> counts of cases, I'm assuming that there can only be 1 a-to-b
> connection at a time. For example if we have a1-b1 connected we
> cannot have a2-b2 at the same time. Thus producing branching chains
> as illustrated in your second link.
>
> Given such labeling, there are a lot more than 360 3-molecule cases.
> Let's consider the case where a has links to b and to c. There are 30
> ways to choose which C,OHs of a to connect from. This is 2*(6 2)
> which is 2*6*5/2 = 30. [Namely, from a1,a2; a2,a1; a1,a3; a3,a1;
> ... a5,a6; a6,a5.] For each of those 30 ways, there are 36 ways to
> connect b and c to a; for example, in the a1,a2 case, we have a1-b1,
> a2-c1; a1-b1, a2-c2; ... a1-b6, a2-c6. Thus, 30*36 = 1080
> arrangements with a-to-b and a-to-c connections.
>
> Now for each of those 1080 arrangements, there are 26 ways to connect
> b-to-c. This is 1+5*5, counting 1 no-b-c-connection case and 25 cases
> which occupy C-OHs that a didn't connect to. 26*1080 = 28080 cases
> with a connected to b and to c, and b-c connected or not.
>
> I haven't worked out the details, but there are a bunch more cases
> where a connects to b but not to c, or where it connects to c but not
> to b. Eg, 36 a-b cases * 30 b-c cases, and 36 a-c cases * 30 c-b
> cases, which if that's all would make the total 28*1080 = 30240.
>
> Anyhow, see if the comments above match up with your chemistry model.
> Thanks! - jiw

If in any given case every molecule (or, "node") connects once each to
all of the others (call this "all-connected") there are easy results
for 1 to 6 nodes.

As an example, consider the 3-node case. With each node connected to
two other nodes, there are 2! * 6C2 = 30 ways (as noted above) to use
each node. The connection points on each node can be set without
regard to what other nodes do, hence the number of all-connected
arrangements is 30^3 = 27000 = 25*1080, consistent with previous
result where not-all-connected configurations added 1*1080 or 3*1080
counts for the 28080 or 30240 counts.

Note, there are m! * 6Cm ways to make m connections from one node.
Now 6Cm = 6!/(m! * (6-m)!) so m!*6Cm = 6!/(6-m)!. Thus, for 1 to 6
nodes, the counts of all-connected arrangements are as below.

1: 0 connections made per node -> 6!/6! = 1 and 1^1 = 1 case

2: 1 conn/node -> 6!/5! = 6 and 6^2 = 36 cases

3: 2 conn/node -> 6!/4! = 30 and 30^3 = 27000 cases

4: 3 conn/node -> 6!/3! = 120 and 120^4 = 207360000 cases

5: 4 conn/node -> 6!/2! = 360 and 360^5 = 6046617600000 cases

6: 5 conn/node -> 6!/1! = 720 and 720^6 = 139314069504000000

On Sunday, April 16, 2023 at 1:58:28 PM UTC-7, James Waldby wrote:
> James Waldby <n...@no.no> wrote:
> > bo reddude <boredd...@gmail.com> wrote:
> >> Let's say you have monomers of glucose(single molecule of glucose).
> > [snip chem stuff and link to 8-point figure]
> >
> >> Each of the 6 points can connect with another identical 6 pointed arrow. If you add more 6 pointed arrows, you get a scenario where the possible combination of configuration increase exponentially.
> >>
> >> For the example of glucose, given 2 molecules of glucose with 6 OHs each, there are 36 possible dimers than can form. (glucose_1 carbon_1-OH can react with 6 OHs on glucose_2. glucose_1 carbon_2-OH can react with 6 OHs on glucose_2 etc for all 6 carbons on glucose_1)
> >>
> >> So given the 6 OHs on gluocse_1, you can react it with 6 OHs on glucose_2
> >>
> >> But if you consider a trimer of glucose, you might think it's just 6*6*6, but it's actually not, because the 3rd glucose can react with 10 remaining OHs on the glucose dimer using its 6 OHs. It's actually 6*6*10.
> >>
> >> And the for the polymer of 4 glucose molecules, the 4th glucose has 15 OHs on the trimer that it can react with using its 6 OHs.
> > [snip link]
> >> What is the formal way to think about this problem using combinatorics for arbitrary length n for the number of glucose molecules in a polymer?
> >
> > Before I think about 4 molecules, let me ask about the 3-molecule
> > case, where you say the count is 6*6*10, or 360. My impression is
> > that you have labeled each molecule (glucose_1, glucose_2, etc but I'm
> > just going to call them a, b, c) and have labeled the carbons and
> > their hydroxyls (carbon_1-OH, etc, which I'll refer to as a1...a6,
> > b1...b6, c1...c6). That is, molecules are distinctly labeled and
> > C-OHs also distinctly labeled; otherwise, instead of there being 36
> > a-to-b cases, there would only be 1 case. Also, from your suggested
> > counts of cases, I'm assuming that there can only be 1 a-to-b
> > connection at a time. For example if we have a1-b1 connected we
> > cannot have a2-b2 at the same time. Thus producing branching chains
> > as illustrated in your second link.
> >
> > Given such labeling, there are a lot more than 360 3-molecule cases.
> > Let's consider the case where a has links to b and to c. There are 30
> > ways to choose which C,OHs of a to connect from. This is 2*(6 2)
> > which is 2*6*5/2 = 30. [Namely, from a1,a2; a2,a1; a1,a3; a3,a1;
> > ... a5,a6; a6,a5.] For each of those 30 ways, there are 36 ways to
> > connect b and c to a; for example, in the a1,a2 case, we have a1-b1,
> > a2-c1; a1-b1, a2-c2; ... a1-b6, a2-c6. Thus, 30*36 = 1080
> > arrangements with a-to-b and a-to-c connections.
> >
> > Now for each of those 1080 arrangements, there are 26 ways to connect
> > b-to-c. This is 1+5*5, counting 1 no-b-c-connection case and 25 cases
> > which occupy C-OHs that a didn't connect to. 26*1080 = 28080 cases
> > with a connected to b and to c, and b-c connected or not.
> >
> > I haven't worked out the details, but there are a bunch more cases
> > where a connects to b but not to c, or where it connects to c but not
> > to b. Eg, 36 a-b cases * 30 b-c cases, and 36 a-c cases * 30 c-b
> > cases, which if that's all would make the total 28*1080 = 30240.
> >
> > Anyhow, see if the comments above match up with your chemistry model.
> > Thanks! - jiw
> If in any given case every molecule (or, "node") connects once each to
> all of the others (call this "all-connected") there are easy results
> for 1 to 6 nodes.
>
> As an example, consider the 3-node case. With each node connected to
> two other nodes, there are 2! * 6C2 = 30 ways (as noted above) to use
> each node. The connection points on each node can be set without
> regard to what other nodes do, hence the number of all-connected
> arrangements is 30^3 = 27000 = 25*1080, consistent with previous
> result where not-all-connected configurations added 1*1080 or 3*1080
> counts for the 28080 or 30240 counts.
>
> Note, there are m! * 6Cm ways to make m connections from one node.
> Now 6Cm = 6!/(m! * (6-m)!) so m!*6Cm = 6!/(6-m)!. Thus, for 1 to 6
> nodes, the counts of all-connected arrangements are as below.
>
> 1: 0 connections made per node -> 6!/6! = 1 and 1^1 = 1 case
>
> 2: 1 conn/node -> 6!/5! = 6 and 6^2 = 36 cases
>
> 3: 2 conn/node -> 6!/4! = 30 and 30^3 = 27000 cases
>
> 4: 3 conn/node -> 6!/3! = 120 and 120^4 = 207360000 cases
>
> 5: 4 conn/node -> 6!/2! = 360 and 360^5 = 6046617600000 cases
>
> 6: 5 conn/node -> 6!/1! = 720 and 720^6 = 139314069504000000

Not sure where the factorial is coming from, but it seems you got the math more or less correct. Thank you. I'm gonna have to take my time going over this.