Genady

Et tu, Brute? The very last time we watched TV was the first Obama inauguration.

In this case, what is a proper label for that kind of social media, the one with the algorithms?

Me too.

The vertical force he exert is equal to the weight. However, the weight itself, i.e. the heavy wheel doesn't stay above his head. If we look only at the vertical displacement of the wheel, it goes up and then down.

I'm quite positive that the OP is talking about this video: No, it doesn't suggest that the weight changes. It is a pretty good demonstration and is followed by explanation.

There is a summation in the formula (2) for A(), but in your calculation there is no summation. So, you have calculated only one components of A(3,2), rather than the value A(3,2). In fact, A(3,2) = 6. For example, if your objects are a, b, and c, then there are these 6 ways to distribute them in 2 distinguishable cells with no empty cells: a | bc b | ac c | ab bc | a ac | b ab | c

No.

A(4-k, 2) depends on k. So, it is not 7. It is different for different values of k.

Which question do you try to answer with this?

The general form is OK, but the variables are not. First, do it the way in which the eq (1) is given. I.e. use A(..., n) as given and derive A(r, n+1). It Should look like A(r, n+1) = .... A(r-k, n). Second, provide explanation for each component, i.e. what is the coefficient in front of A(r-k, n) and what values of k are you summing for.

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You need to do it algebraically, not numerically. Did you see my suggestion about how to do it? Did you try it?

I've given you my suggestions in my previous post.

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1. You need to derive the equations for generic variables r and n. Working with specific values, such as r=4 and n=2, will not give you a general derivation. 2. The upper limit of summation is still wrong. 3. I don't know where the second part of these two simple sums came from. From nowhere? 4. You still did not derive equation (1). I've given you a fat hint for that. If you derive it, you will see where your summation limit is wrong.

Here you go:

It is not the order of summation you need to change. It is the upper limit of summation.

I think, the difference appears because of the problem which I've pointed to earlier:

Yes, both equations (1) and (2) relate to distinguishable objects and distinguishable cells. Eq (1) is recursive. That is, assume that there are A(x, n) ways to distribute x distinguishable objects between n distinguishable cells, with no empty cells. Knowing this for any number of distinguishable objects x > n, we want to find A(r, n+1), number of ways to distribute r distinguishable objects between n+1 distinguishable cells, with no empty cells. Start thinking about it in the following way. Take one of the n+1 cells aside. Let's call it a "new" cell. You have one new cell and n old cells. You have to put some number of objects, k>0, in the new cell. The rest r-k objects will go into the n old cells. This can be done in A(r-k, n) ways, the number that we assume we know. Can you continue from here so we get the expression for A(r, n+1), which is the eq (1)?

In the eq (1) here your upper limit of summation is k=r. Are you sure about it? I don't think it is correct. The case (2) doesn't make sense. You cannot distribute 2 objects in 3 cells such that there are no empty cells. I think, you have to have at least as many objects as there are cells for these formulas to make sense. That's why I think that the upper limit of summation in eq (1) is incorrect. Did you derive the eq (1) by a combinatorial argument, as requested?

You can substitute A(r-k, n) in eq.(1) using the right side of eq.(2) while instead of (n-v)r you should put (n-v)r-k. Let's see what you get.

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Yes, this formula is correct. However, the equation (2) is not. Take for example r=2 objects and n=2 cells. There is only 1 way to distribute them with no empty cells, i.e. 1+1. But the equation (2) gives 2. This would be so if, for example, the objects were distinguishable.

OK. Now, the first formula in the OP says, A(r,n) = C(r-1,n-1) Where did it come from?

Here is your other post: Of course v or k doesn't make a difference. There is an actual difference between (2) above and the second formula in the OP: Do you see it? Plus, I don't see anything there about the equivalence you're talking about here.