Timeline for Combinatorial equation
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 24, 2010 at 3:17 | vote | accept | Amir | ||
| Oct 24, 2010 at 3:16 | vote | accept | Amir | ||
| Oct 24, 2010 at 3:17 | |||||
| Oct 24, 2010 at 3:15 | vote | accept | Amir | ||
| Oct 24, 2010 at 3:16 | |||||
| Oct 24, 2010 at 2:54 | comment | added | user6976 | @Richard: Is there a "1:1 correspondence" type proof of it. Say, if you, indeed, represent $n^n$ as the size of the set of all functions $\{1,..,n\}$ to itself? | |
| Oct 24, 2010 at 2:46 | comment | added | Richard Stanley | The identity is a special case of a class of identities known as "Abel's (binomial) identity." See for instance mathworld.wolfram.com/BinomialIdentity.html. For a more general result, see Exercise 5.31 of my book Enumerative Combinatorics, vol. 2. | |
| Oct 24, 2010 at 2:37 | comment | added | Amir | Thanks for your answer. Actually, I did not want to use Lagrangian inversion formula!. I was hoping if one can give me a combinatorial proof for this. For example, n^n is the number of functions from {1..n} to {1..n} or it can be considered as the number of strings with length n over alphabet {1..n}. But I could not find a way to count the same objects using the right-hand side of the equation. | |
| Oct 24, 2010 at 2:24 | history | edited | Matti Åstrand | CC BY-SA 2.5 |
display math
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| Oct 24, 2010 at 2:17 | history | answered | Matti Åstrand | CC BY-SA 2.5 |