jason
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 Feb13 awarded Popular Question May3 accepted Stirling Numbers of the Second Kind (deriving S(m, m - 2)) May3 comment Stirling Numbers of the Second Kind (deriving S(m, m - 2)) Thanks for that, dividing by 2 gave me your expected result. I checked this with a few test numbers and it seems like the book made a mistake. :) May3 comment Stirling Numbers of the Second Kind (deriving S(m, m - 2)) The book defined Stirling numbers of the second kind as: S(m, n) = 1/n! * sum(k=0 to n)[((-1)^k) * $\dbinom{n}{n-k}$(n - k)^m] Sorry about the formatting of it, I'm not really familiar with LaTeX May3 asked Stirling Numbers of the Second Kind (deriving S(m, m - 2)) Apr25 accepted Combinations with Repetition, bounded above and by a subset Apr24 revised Combinations with Repetition, bounded above and by a subset fixed spelling mistake Apr24 asked Combinations with Repetition, bounded above and by a subset Apr24 awarded Scholar Apr24 accepted 20 books 5 different shelves Apr24 comment 20 books 5 different shelves Ahh, that clears it up. Thanks a lot :) Is there a way to mark your answer is the correct one? Sorry, I'm new here. Edit: Found it Apr24 comment 20 books 5 different shelves Thanks for the reply, I was wondering if you could clarify the first part where you say there are 15^6 ways to place the books. I guess I'm having a mind blank, because I can't really see how you got there. Once one book is put onto a shelf, the remaining books decrease so I thought this would lead to a factorial of some kind. Thanks Apr24 awarded Student Apr24 asked 20 books 5 different shelves