I came across the following question and it's a bit different from what I'm used to...

Write a generating function for each of the following:

1) You are making an Easter basket with at most two special edition Peeps. Each special edition Peep comes in its own one candy package. There are 4 different types of special edition Peep's: red velvet, lemon spice, party cake, and lemonade. If you have 2 peeps, they could either be the same or different flavors.

2) You are making an Easter basket and filling it with n pieces of candy. You can include any number of packages of 5 yellow chick Peeps, any number of packages of 5 blue bunny Peeps, and special edition peeps following the requirements in the previous problem.


For (1), I'm being thrown off by the "at most two" requirement. Usually generating functions deal with infinite series with an arbitrary large n. At first I thought maybe the following might be it:


But I'm not sure at all. If I were allowed to express it as an integer addition question, I thought maybe I could express it like so:

$$x_1+x_2+x_3+x_4 \leq 2, x_{1,2,3,4} \geq 0$$ which can be solved via the stars & bars method: ${6 \choose 4}=15$ (is this right?) If so, the above generating function doesn't result in the coefficient of 15 when n = 2. I feel like I'm just thinking about this wrong.

For (2), since I'm stuck on (1) I don't know where to start on this.

Thanks for any help.


EDIT2: So it looks like I'm getting close on (1)...

Also, for (2), is the first part of it just $(1+x+x^2+x^3+x^4+x^5)^2$ or am I missing something? Thanks.

  • $\begingroup$ Oh wait, since n is at most 2, am I supposed to take into consideration all values of n from 0 to 2? If so the above generating function does result in the sum of 15... $\endgroup$ – user3280193 Apr 19 '16 at 2:33
  • $\begingroup$ Your comment is correct; you’ve now almost answered the first question. If $g(x)=(1+x+x^2)^4$, you want the function such that the coefficient of $x^n$ is the sum of the coefficients in $g$ of $x^k$ with $k\le n$. Do you know what to multiply $g(x)$ by to get that? $\endgroup$ – Brian M. Scott Apr 19 '16 at 2:34
  • $\begingroup$ Hmm...I guess I'm stuck again. :D I've just never run into this type of question variant before. $\endgroup$ – user3280193 Apr 19 '16 at 2:43
  • $\begingroup$ Useful fact (that you should try to prove): if $g(x)=\sum_{n\ge 0}a_nx^n$, then $$\frac1{1-x}\cdot g(x)=\sum_{n\ge 0}\left(\sum_{k=0}^na_k\right)x^n\;.$$ Multiplying by $\frac1{1-x}$ gives you the partial sums of the sequence of coefficients of the original series. $\endgroup$ – Brian M. Scott Apr 19 '16 at 2:45
  • $\begingroup$ Awesome, I'll think about this. Thanks. $\endgroup$ – user3280193 Apr 19 '16 at 3:06

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