A Combinatorical Proof

Prove the following combinatorial:

The number of ways of placing $n$ labeled balls into $n$ indistinguishable boxes is equal to

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

#Combinatorics

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 4 years, 9 months ago

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

Sharky Kesa - 4 years, 9 months ago

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 - 4 years, 9 months ago

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

Sharky Kesa - 4 years, 9 months ago

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

Wen Z - 4 years, 9 months ago

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

Sharky Kesa - 4 years, 9 months ago

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 - 4 years, 9 months ago

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

Sharky Kesa - 4 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$