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It is my understanding that we tend to use ordinary generating functions for indistinguishable sets of items & exponential generating functions for the opposite. While I have this baseline knowledge, I've never been shown how exponential generating functions help with this where ordinary generating functions cannot.

To my knowledge, an ordinary generating function helps with counting simply due to the convenience of the power series produced by $\frac{1}{1-x}$. I don't quite see how the equation we use for the exponential generating function is any different. Given $e^nx = \sum_{k=0}^{\infty}\frac{n^k}{k!}x^k$.

Could somebody please give me some insight into why this is?

ryno
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    https://math.stackexchange.com/questions/4371476/permutations-of-3-items-with-duplication-where-pqr-n has an example involving getting multiple factorials into the denominator but needing to multiply by a factorial to get the final result, essentially the use of exponential generating functions since the final step could not be expressed with ordinary generating functions – Henry Feb 17 '22 at 22:21
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    It's because the multiplication of exponential generating functions corresponds exactly to the natural way to combine together labeled structures (i.e. distinguishable sets). I went into more detail in this answer. – Mike Earnest Feb 17 '22 at 23:31
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    Among other things, using exponential generating functions lets us find a closed form for generating functions of sequences where a closed form for the ordinary generating functions just does not exist. For example, consider the ogf for the ${n!}{n=0}^{\infty}$ sequence, $\sum{n=0}^{\infty}n!x^n$. Its radius of convergence is $0$. There is only so much you can do with it, even formally. But consider the egf for the same sequence, $\sum_{n=0}^{\infty}n!\frac{x^n}{n!}=\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$. Now that's a lot easier to handle, it has a closed form, and so is a lot more useful. – Alexander Burstein Feb 18 '22 at 06:21

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