# Problem 06

## Problem 06

If two classical dices are thrown, the sum has generating function

F(x) = x2 + 2x3 + 3x4 + 4x5 + 5x6 + 6x7 + 5x8 + 4x9 + 3x10 + 2x11 + x12 = x2 (1 + x)2 (1 + x + x2)2 (1 - x + x2)2.

One can write other positive integers on the 12 faces to obtain the same sum distribution. Find the (unique) other way.

### Solution

We are looking for a1, ..., a6, b1, ..., b6, such that F(x) = (xa1 + xa2 + ... + xa6)(xb1 + xb2 + ... + xb6) = G(x)H(x).

Since we need positive integers, x must divide both factors. Since G(1) = H(1) = 6, the factors 1 + x and 1 + x + x2 must also appear exactly once in both G(x) and H(x). Hence, the only choice is whether to put the two factors 1 - x + x2 together or not.

The first option yields G(x) = H(x) = x (1 + x) (1 + x + x2) (1 - x + x2) = x1 + x2 + x3 + x4 + x5 + x6, which yields the classical dice. The second option yields the unique other solution; one dice with {1, 3, 4, 5, 6, 8} and one dice with {1, 2, 2, 3, 3, 4}:

G(x) = x (1 + x) (1 + x + x2) (1 - x + x2)2 = x1 + x3 + x4 + x5 + x6 + x8,
H(x) = x (1 + x) (1 + x + x2) = x1 + x2 + x2 + x3 + x3 + x4,

### Remarks

Students often wonder why they should study generating functions. Artificial counting problems don't convince them. And I can't blame them. Alas, most routine exercises on this subject are just counting problems. This is a different kind of problem, with for most an unexpected result. I googled this and it turns out to be known as the Sicherman dice.