How to eliminate the repeated case from polynomial counting? How to eliminate the repeated case from polynomial counting?
assume a die throw $3$ times, do not allow repeated number appear
$$(x+x^2+x^3+x^4+x^5+x^6)^3 - y$$
how to count the repeated case, that should be minused in above polynomial counting?
what is $y$ in terms of polynomial of $x$?
Mark six maple code
subs(z=0, diff(expand((1+z*x)*(1+z*x^2)*(1+z*x^3)*(1+z*x^4)*(1+z*x^5)*(1+z*x^6)*(1+z*x^7)*(1+z*x^8)*(1+z*x^9)*(1+z*x^10)
*(1+z*x^11)*(1+z*x^12)*(1+z*x^13)*(1+z*x^14)*(1+z*x^15)*(1+z*x^16)*(1+z*x^17)*(1+z*x^18)*(1+z*x^19)*(1+z*x^20)
*(1+z*x^21)*(1+z*x^22)*(1+z*x^23)*(1+z*x^24)*(1+z*x^25)*(1+z*x^26)*(1+z*x^27)*(1+z*x^28)*(1+z*x^29)*(1+z*x^30)
*(1+z*x^31)*(1+z*x^32)*(1+z*x^33)*(1+z*x^34)*(1+z*x^35)*(1+z*x^36)*(1+z*x^37)*(1+z*x^38)*(1+z*x^39)*(1+z*x^40)
*(1+z*x^41)*(1+z*x^42)*(1+z*x^43)*(1+z*x^44)*(1+z*x^45)*(1+z*x^46)*(1+z*x^47)*(1+z*x^48)*(1+z*x^49)), z$6));


Instead of subtracting out the repeated cases, it's easier to only generate the non-repeated cases in the first place:
$$(1+zx)(1+zx^2)(1+zx^3)(1+zx^4)(1+zx^5)(1+zx^6)\;.$$
Then the coefficient of $z^3x^n$ counts the number of partitions of $n$ into $3$ distinct parts from $1$ to $6$, and the number of ways of getting that sum from the dice is $3!$ times that number, which is the coefficient of $x^n$ in
$$\left.\frac{\partial^3}{\partial z^3}\left((1+zx)(1+zx^2)(1+zx^3)(1+zx^4)(1+zx^5)(1+zx^6)\right)\right|_{z=0}\;.$$
