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I'm trying to figure out the rule that's producing the series below. The first column is the factorials $1!,2!,3!,4!,5!,6!$ But I can't figure out what else is going on. So I know the next row will start with $7!=5040$, but can't say much beyond that. If someone has insight as to what the next numbers in the series are that would be much appreciated.








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Have you tried dividing each row by the first term? – Qiaochu Yuan Jan 3 '12 at 7:08
3… – sdcvvc Jan 3 '12 at 7:08
@QiaochuYuan Commented while I was posting my answer, d'oh! – Steven Stadnicki Jan 3 '12 at 7:12
@sdcvvc what a cool website – ben Jan 3 '12 at 7:29
up vote 5 down vote accepted

The easiest way to find the answer here is to divide out by the first terms, producing:

$$\begin{array} &&&&&&1\\ &&&&1&&1\\ &&&1&&2&&1\\ &&1&&3&&3&&1\\ &1&&4&&6&&4&&1\\ \end{array}$$ ...which should look awfully familiar. This implies that the $(n,k)$ entry of your series is $n! {n-1 \choose k}$.

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Thanks! Much clearer now. Pardon my ignorance but I am not familiar with the notation you used. What exactly does $n! {n-1 \choose k}$ mean? – ben Jan 3 '12 at 7:32
Well, the $n!$ piece of that you already know; $n-1\choose k$ (generally pronounced '$n$-minus-1 choose $k$') is the Binomial Coefficient. $a\choose b$ is defined as $a!/b!(a-b)!$ and represents the number of ways of choosing b items from a population of a items; they're arguably the most fundamental objects in combinatorics. See for more details on it! – Steven Stadnicki Jan 3 '12 at 7:56

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