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A Combinatorical Proof

Prove the following combinatorial:

The number of ways of placing \(n\) labeled balls into \(n\) indistinguishable boxes is equal to

\[\dfrac{\displaystyle \sum_{k=0}^{\infty} \dfrac {(k+1)^{n-1}}{k!}}{e}\]

Note by Sharky Kesa
1 year, 3 months ago

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What you are counting is the Bell numbers, and your result is known as Dobinski's formula.

(By the way, this is not a bijection. A bijection is a 1-1 correspondence between two sets. What you want to prove is that a certain combinatorial formula is true.)

Jon Haussmann - 1 year, 2 months ago

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OK, thanks! However, I thought that a bijection could be constructed between these two by changing the summation.

Sharky Kesa - 1 year, 2 months ago

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Okay so I have a solution for putting \(n\) balls in exactly \(k\) boxes:

Define a function that takes an ordered pair of naturals to a nonnegative integer:

\(f(n,k)=\begin{cases} 1 & k=1 \\ 1 & n=k \\0 & k>n\\kf(n-1,k)+f(n-1,k-1) & \text{for other cases}\end{cases}\)

This function should satisfy it if my LaTeX is correct.

Now for putting \(n\) balls in \(n\) boxes, just do

\(\displaystyle \sum_{x=1}^{n}f(n,x)\)

Wen Z - 1 year, 3 months ago

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Now you just need to biject this summation to the other summation.

Sharky Kesa - 1 year, 3 months ago

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I looked it up and it I haven't seen any solutions which use a combinatorial bijection.

Wen Z - 1 year, 3 months ago

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@Wen Z Can you think of any other solution tghen? It might help to create a bijection.

Sharky Kesa - 1 year, 3 months ago

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Are you sure that's correct? For example when \(n=1\) we have \(1\) way of putting a ball in a box, but the sum indicates that there are a few more than 1 way; for \(k=0\) that term alone is already \(1\).

Wen Z - 1 year, 3 months ago

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I forgot to put a denominator in the sum but it has been rectified.

Sharky Kesa - 1 year, 3 months ago

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