Prove the following combinatorial:

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

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

Prove the following combinatorial:

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

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

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TopNewestWhat 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 month ago

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– Sharky Kesa · 1 month ago

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

Okay so I have a solution for putting \(n\) balls in

\(k\) boxes:exactlyDefine 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 month ago

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– Sharky Kesa · 1 month ago

Now you just need to biject this summation to the other summation.Log in to reply

– Wen Z · 1 month ago

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

– Sharky Kesa · 1 month ago

Can you think of any other solution tghen? It might help to create a bijection.Log in to reply

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 month ago

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– Sharky Kesa · 1 month ago

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