An ordinary enumerator is given as $(1+x+x^2)^p$. This is being understood as follows:

There are 2 each of p kinds of objects.The ordinary enumerator for selecting none (or) one (or) both the objects of that kind is $(1+x+x^2)^p$

Similarly,how do i visualize (or) understand the enumerator
$(1+x^5+x^9)^{100}$ in plain english

  • $\begingroup$ It seems to me the example "plain English" statement relies on the term ordinary enumerator without giving a definition. At best the statement is a kind of illustrative example, but it doesn't seem all that clear (perhaps leading to your difficulty in applying it to a quite similar example). Should we back up and give an understandable account of generating functions? $\endgroup$ – hardmath Jun 17 '15 at 15:06
  • $\begingroup$ @hardmath The actual question was to find coeffecient of (x^23) in (1+x^5+x^9)^100.An explanation leading to it's answer would be really great. $\endgroup$ – Pradeep Jun 17 '15 at 15:12
  • $\begingroup$ We could adapt the methods of computation I used in this recent Answer to find that coefficient. The wording of the Question above suggests you want an interpretation of (all) coefficients, which probably has a large number of valid responses, while the algebra of the polynomial, though somewhat tedious, has a well-defined result. $\endgroup$ – hardmath Jun 17 '15 at 15:20
  • $\begingroup$ @hardmath In the link mentioned above, how did we convert $(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2$ to $(\frac{1 - x^{21}}{1 - x})^2$. Is there a formula ? $\endgroup$ – Pradeep Jun 17 '15 at 15:42
  • $\begingroup$ This is just the usual formula for a geometric sum. It doesn't apply in any obvious way to your problem. $\endgroup$ – hardmath Jun 17 '15 at 15:45

Relating that polynomial to your own example, we can understand $(1+x^5+x^9)^{100}$ as follows:

There are $9$ each of $100$ kinds of objects. The ordinary enumerator for selecting none or five or all nine of the objects of that kind is $(1+x^5+x^9)^{100}$.

Here's another interpretation:

Consider three-sided dice whose faces have $0$, $5$, or $9$ pips, and roll $100$ such dice. Then the coefficient of $x^k$ in $(1+x^5+x^9)^{100}$ is the number of ways to get a total of $k$.

  • 1
    $\begingroup$ The second example was great.Thanks..!! $\endgroup$ – Pradeep Jun 17 '15 at 16:51
  • $\begingroup$ @Pradeep I find the second example a lot easier to understand too! $\endgroup$ – Théophile Jun 17 '15 at 18:07

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