There are three $4-$faced black distinct dice whose faces are numbered from $2$ to $5$ and three $4-$faced white distinct dice whose faces are not numbered.

Let $a_n$ be the number of ways in tossing the $3$ black dice and get the total of $n$ and $b_n$ be the number of ways in tossing the $3$ white dice and get the total of $n$.

In how many ways can we number the faces of the $3$ white dice so that $a_n = b_n$ for all $n \in \mathbb{N}$ ?

I just know how to find $a_n$ by using Generating function but I can't figure out how to solve the problem. Thank you.

  • $\begingroup$ So, you need to look for non-standard partitions of the factored form of the generating function. See Sicherman Dice. Here your generating function is $x^6(x+1)^3(x^2+1)^3$. You need four faces so in each partition the coefficients add to $4$. I assume you only want positive integer solutions? $\endgroup$ – lulu Dec 31 '16 at 12:59
  • $\begingroup$ My generating function for black dice is $ x^6 \frac{(1-x^5)^3}{(1-x)^3} $ $\endgroup$ – carat Dec 31 '16 at 14:54
  • $\begingroup$ It's $(x^2+x^3+x^4+x^5)^3=x^6(1+x+x^2+x^3)^3=x^6(x+1)^3(x^2+1)^3$. You need to work with the factors so the quotient form you use, while correct, isn't the most helpful. $\endgroup$ – lulu Dec 31 '16 at 15:05
  • $\begingroup$ Sorry, the numerator of the fraction should be $(1-x^4)^3$. $\endgroup$ – carat Dec 31 '16 at 15:17
  • $\begingroup$ Right. But, as I say, the factored form is what you want here. $\endgroup$ – lulu Dec 31 '16 at 15:19

Consider probability generating function for rolling one die,

$A(x) = \frac{1}{4}(x^2 + x^3 + x^4 + x^5)$ $4A(x) = x^2 + x^3 + x^4 + x^5 = x^2(1+ x+x^2 + x^3)= x^2(1+ x)(1+x^2)$

Probability GF for rolling three dice is,

$4^3\cdot A^3(x) = x^6(1+ x)^3(1+x^2)^3$

We'll partition the factors on RHS to form 3 white dice.

Each partition must contain $x$ as the number on each face is at least one.

Since LHS contains factor 4, so GF for each partition must have two factor 2.

As $\Phi_2(x) = 2, \Phi_4(x) = 2$, so $(1+ x)^3(1+x^2)^3$ can be partitioned in 2 ways,


$(1+ x)(1+x^2)$

$(1+ x)(1+x^2)$

$(1+ x)(1+x^2)$

and $x^6$ can be distributed as follow,

4 1 1

3 2 1

2 2 2 <-- cancelled as this would be the same as original black dice.

Number of ways = 2

2 :

$(1+ x)^2$


$(1+ x)(1+x^2)$

Number of ways to distribute $x^6$ = $\binom{r-1}{n-1} = \binom{6-1}{3-1} = \binom{5}{2} = 10$

Ans : total number of ways = 10 + 2 = 12


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