Dgdrxrt

Consider three sets of cards colored Blue, Red and Yellow. Each set has cards numbered $1-10$. The $4$ remaining cards are all indistinguishable cards numbered $0$.

Card numbered $i$ has the value of $2^i$. How many ways are there to choose a group of cards that sums up to $2016$.

I'm having a problem with creating a generating function for this problem, any help would be appreciated.

So far I have $f(x) = (x^1 + x^2 + x^3 + x^4)cdot(1 + x^2^1 + x^2^2 + x^2^3 ... + x^2^10)^3$ but I'm not really sure that's even correct.

We can represent the number of selecting zero up to four indistinguishible cards which have value $2^0=1$ as
beginalign*
1+x+x^2+x^3+x^4=frac1-x^51-x
endalign*

We can represent the number of selecting zero up to three distinguishible cards numbered with value $2^i$ as
beginalign*
left(1+x^2^iright)^3qquad 1leq ileq 10
endalign*

The corresponding generating function is
beginalign*
&frac1-x^51-xleft(1+x^2^1right)^3left(1+x^2^2right)^3left(1+x^2^3right)^3cdotsleft(1+x^2^10right)^3\
&qquad=frac1-x^51-xleft(sum_j=0^2^10-1x^2jright)^3\
&qquad=frac1-x^51-xleft(frac1-x^2^111-x^2right)^3tag1
endalign*
in accordance with the generating function given by the answer from @RossMillikan.

Next we have to extract the coefficient of $x^2016$ from (1). It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series.

We obtain from (1)
beginalign*
colorblue[x^2016]&colorbluefrac1-x^51-xleft(frac1-x^2^111-x^2right)^3\
&=[x^2016]frac1-x^51-xleft(frac11-x^2right)^3tag2\
&=left([x^2016]-[x^2011]right)sum_j=0^infty x^ksum_j=0^inftybinom-3j(-x^2)^jtag3\
&=left(sum_k=0^2016[x^k]-sum_k=0^2011[x^k]right)sum_j=0^inftybinomj+22x^2jtag4\
&=left(sum_k=0^1008[x^2k]-sum_k=0^1005[x^2k]right)sum_j=0^inftybinomj+22x^2jtag5\
&=binom10082+binom10092+binom10102tag6\
&,,colorblue=1,525,609
endalign*

Comment:

• In (2) we omit terms in the numerator with powers of $x^2^11$ since they do not contribute to the coefficient of $x^2016$.




• In (3) we apply $[x^p-q]A(x)=[x^p]x^qA(x)$ and do a geometric and binomial series expansion.




• In (4) we use the binomial identity $binom-pq=binomp+q-1p-1(-1)^q$.




• In (5) we skip coefficients of odd powers since they do not contribute.




• In (6) we select the coefficients accordingly.

Here are a couple of hints:

With the blue cards, you can make any even number from $0$ to $2^11-2$, and you can make it it exactly one way. (Think binary numbers.)

With the the zero cards, you can make $0$ in $1$ way, $1$ in $4$ ways, $2$ in $6$ ways, etc. if we take account of which zero card we use. From the statement that the zero cards are indistinguishable, however, I suppose we can make $0,1,2,3,$ or $4$ in one way.

Can you finish it from here?

The blue cards can give you any even value from $0$ through $2046$ in exactly one way, so their generating function is $1+x^2+x^4+ldots+x^2046=frac 1-x^20481-x^2$. You could also get here by saying the $1$ card has value $2$ so generating function $1+x^2$, the $2$ card has function $1+x^4$ and so on. Multiply all those together for a given color and you get the above. Each color can give you the same, so you cube it. The blanks can give you any value from $0$ to $4$, so their generating function is $1+x+x^2+x^3+x^4=frac 1-x^51-x$ and the overall generating function is
$$left(frac 1-x^51-xright)left(frac 1-x^20481-x^2right)^3$$