[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 467

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PapaYozh:

I'm slowing down, give me a formula.

---

There is a set of characters. The number of characters is 2 * N, i.e. even.

Characters are divided into 2 subsets of N characters each. Determine the number of possible ways to divide symbols into subsets. The position of the symbol in the subset is not important.

That is:

1) For set {A,B} (i.e. with N=1) there is a single division option: {A} + {B}

2) For the set {A,B,C,D} (i.e. for N=2) there are 3 variants:

{AB} + {CD}

{AC} + {BD}

{AD} + {BC}

---

How to determine the number of variants for an arbitrary value of N ?


It turns out that the formula would be = n!/(number of elements of the two sets)/2. Well that just caught my eye. The letters are 4 - ABCD. That is 4 elements. 4! = 24. 24/4=6 6/2=3

You have to check, though. So for number 6 the number of combinations will be by this formula = 6!/6/2 = 60 combinations.

 
drknn:


It turns out that the formula would be = n!/(number of elements of two sets)/2. Well it just caught my eye. The letters are 4 - ABCD. That is 4 elements. 4! = 24. 24/4=6 6/2=3

Although, it has to be checked. So for number 6 the number of combinations will be according to this formula = 6!/6/2 = 60 combinations.


For 6, I get 38 options.
 
PapaYozh:

For 6, I get 38 options.

6! = 6*5*4*3*2*1 = 30*4*3*2 = 120*3*2 = 360*2 = 720. 720/6 = 120 120/2 = 60
 
Wait, something is wrong here, PapaYozh, do you really need this, or is it just another simulator? I'm asking you: should I pick up my Combinatorics textbook, or can I forget about this problem? There's a formula somewhere in the outline. There is a fixed problem of this type.
 
PapaYozh:

I'm slowing down, give me a formula.

---

There is a set of characters. The number of characters is 2 * N, i.e. even.

Characters are divided into 2 subsets of N characters each. Determine the number of possible ways to divide symbols into subsets. The position of the symbol in the subset is not important.

That is:

1) For set {A,B} (i.e. with N=1) there is a single division option: {A} + {B}

2) For the set {A,B,C,D} (i.e. for N=2) there are 3 variants:

{AB} + {CD}

{AC} + {BD}

{AD} + {BC}

---

How to determine the number of choices for an arbitrary value of N ?

The number of ways to divide the set in the given way is exactly 2 times less than the number of ways to choose N symbols from 2*N (not considering the order), because choosing one of the halves is equal to choosing the other. Then by definition the number of ways is equal to the number of combinations of 2N by N divided by 2

i.e. X = 1/2 * C (2N,N) = 1/2 * (2N)!/(N!*(2N-N)!).

For the case N=2 we have X = 1/2 * C(4,2) = 1/2 * 4!/(2!*2!) = 3

For N=3 we have X = 1/2 * C(6,3) = 1/2 * 6!/(3!*3!) = 10

For N=4 X = 1/2 * C(8,4) = 1/2 * 8!/(4!*4!) = 35

For N=5 X = 1/2 * C(10,5) = 1/2 * 10!/(5!*5!) = 126

For N=6 X = 1/2 * C(12,6) = 1/2 * 12!/(6!*6!) = 462

Characteristically, 38 does not work anywhere...

 
alsu:

The number of ways to divide the set in the given way is exactly 2 times less than the number of ways to choose N characters from 2*N (not considering the order), because choosing one half is equal to choosing the other half. Then by definition the number of ways is equal to the number of combinations of 2N by N divided by 2

i.e. X = 1/2 * C(2N,N) = 1/2 * (2N)!/(N!*(2N-N)!).

For the case N=2 we have X = 1/2 * C(4,2) = 1/2 * 4!/(2!*2!) = 3

For N=3 we have X = 1/2 * C(6,3) = 1/2 * 6!/(3!*3!) = 10

For N=4 X = 1/2 * C(8,4) = 1/2 * 8!/(4!*4!) = 35

For N=5 X = 1/2 * C(10,5) = 1/2 * 10!/(5!*5!) = 126

For N=6 X = 1/2 * C(12,6) = 1/2 * 12!/(6!*6!) = 462

Characteristically, 38 does not work anywhere...

Thanks.

Apparently, I was in a hurry.

 
Richie:

Now for the reaction rate. Well, of course, the reaction rate depends on the temperature. The higher it is, the higher it is.


So it seems to be the exact opposite of what Mathemat was saying:

Now we're going to cool the combustion zone. In other words, we will remove the heat. The reaction, according to the Le Chatelier principle, will align the equilibrium so as to minimise the external influence (heat removal). It will 'tend' to generate more heat. Since we have heat on the right side of the reaction, the equilibrium will shift to the right. The fire will intensify.

Or is it my humanitarian brain that's not getting it?
 
Cod:


It seems to be the exact opposite of what Mathemat said:

Now we are going to cool down the combustion zone. In other words, we will remove the heat. The reaction, according to Le Chatelier's principle, will align the equilibrium so that the external influence (heat removal) is minimised. It will 'tend' to generate more heat. Since we have heat on the right side of the reaction, the equilibrium will shift to the right. The fire will intensify.

Or is it my humanitarian brain that doesn't get it?

There is something wrong with this principle. According to it: "The worse, the better!", i.e. no matter how bad the conditions get, the result gets better and better.

:)

 
PapaYozh:

There is something wrong with this principle. According to it: "The worse, the better!", i.e. no matter how bad the conditions get, the result keeps getting better and better.

:)


Well, I'm purely humanitarian (I'm ashamed, but in the early 90's the block was not up to the choice of the university) here's where I hesitated: "The reaction, according to le Chatelier's principle, will align the equilibrium so as to minimise external influence"... Philosophically, how do the atoms involved in the reaction know what is an "external" influence and what is an "internal" one? It's just a point of view, nicht war? Or am I somewhere terribly stupid?
 
Richie:

First of all, let's start with oxygen. Oxygen is obtained by distillation of air. Although there are more modern technologies - membrane technology, for example, but it is not at industrial scale. There is no need to heat it. It will heat itself in an "autogene".

I take it that the useful information for the task is the one that is not in small print. The question wasn't about what made sense, but what would happen to the response rate.

Now about Na.Who says that sodium can't be mixed with water without reacting? You can if water and sodium are solids. In solid form, they do not react with each other. Think of effervescent aspirin tablets where acetylsalicylic acid and citric acid are mixed with sodium bicarbonate. Once in water - reaction, in dry form - no reaction.

No comment: I didn't mix sodium with ice.

Now for the reaction rate. Well, of course, the rate of reaction depends on the temperature. The higher it is, the higher it is. But let's remember the physical chemistry. What else does it depend on? Concentration. And concentration depends on what? Density, for example. I'm talking about gases. Density, by the way, is inversely proportional to temperature. So, as temperature increases from this point of view, concentration of initial substances decreases.

One more point. The reaction rate depends on the concentration of combustion products. The higher concentration of combustion products is, the lower reaction rate is.

So, the question is not very "linear". And it will be "blurted out" on this forum and there will be no definite answer.

As for me, I can't give an unequivocal answer. On the one hand I know that the reaction rate increases with increasing temperature, on the other hand it falls (and there are examples of its use in engineering, particularly in space). The combustion process is 'auto-balanced'. That's why we're all here and not there yet .....

Not everything is right here. But the "highlighted" in regular font is exactly what illustrates the le Chatelier principle.

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