# 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));  - add comment ## 1 Answer 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

$$\left.\frac{\partial^3}{\partial z^3}\left((1+zx)(1+zx^2)(1+zx^3)(1+zx^4)(1+zx^5)(1+zx^6)\right)\right|_{z=0}\;.$$

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very useful for counting mark six gambling –  M-Askman Mar 3 '12 at 13:20
after diff and subs z = 0, the whole equation become zero –  M-Askman Mar 4 '12 at 9:51
@Marco: I don't see any equations here. In case you mean the expression I wrote down, please explain why you believe its value is zero. Wolfram|Alpha thinks otherwise. –  joriki Mar 4 '12 at 15:34
sorry it is not zero, 6*x^15+6*x^14+12*x^13+18*x^12+18*x^11+18*x^10+18*x^9+12*x^8+6*x^7+6*x^6, if there is no ordering, i think that no need to multiply 3!, thank you –  M-Askman Mar 5 '12 at 1:39
@Marco: Your question failed to specify whether you wanted to distinguish by order; I inferred that you did from the facts that the expression you gave for the count with repeated cases distinguished by order and that in the context of dice rolls one is usually interested in probabilities, for which order needs to be distinguished. –  joriki Mar 5 '12 at 4:52