How can I write $n!$ using $\sum$?
Should I write $\sum\limits_{k=2}^n k$?
The domain is $n>1$
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do you know what $n!$ means? – Just dropped in May 10 '18 at 17:22
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5Maybe $n! = \exp\left(\sum_{k=1}^n \log k \right)$ – angryavian May 10 '18 at 17:22
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@JustDroppedIn It's the factorial – user557276 May 10 '18 at 17:22
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Are you sure you don't mean $\Pi$ for product? – Kevin Long May 10 '18 at 17:23
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1May be $$n!=\sum_{k=1}^{n}(n-1)!$$ – Bumblebee May 10 '18 at 17:51
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You can also do it by partitioning the permutation count in terms of number of fixed points. – Randall May 11 '18 at 04:05
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$$n! = \sum_{k=1}^n\left[n\atop k\right]$$ – sku May 12 '18 at 20:43
2 Answers
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Note that for non-negative integer $n$, $$ n! = 1 \times 2 \times \ldots \times n = \prod_{k=1}^n k. $$ If you want to write $n!$ using a sum-like expression, note that $$ \ln(n!) = \ln\left(\prod_{k=1}^n k \right) = \sum_{k=1}^n \ln k, $$ so $$ n! = \exp\left(\sum_{k=1}^n \ln k\right), $$ but not sure this is what you are looking for.
gt6989b
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$n!$ involves the multiplication of subsequent terms, rather than an addition, and therefore it is a product rather than a sum. Therefore, we would typically use product notation, signified by $\prod$ as opposed to $\sum$.
The best expression is $$\prod_{k=1}^{n}{k}$$
Rhys Hughes
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