A die consists of six faces with each face representing precisely one of the numbers 1, 2, 3, 4, 5, 6. Suppose that *n* such dice are rolled for some positive integer *n*. The number of the upper face of each dies is noted and the sum of these numbers are computed. Show that *the number of ways to get an even sum* is

[tex] \displaystyle \sum_{m=0}^N H^3_{2m} H^3_{n-2m} [/tex]

where [tex] H^r_{s} = \binom{r+s-1}{s}[/tex] and N is the largest integer such that [tex]2N \le n[/tex].

Remark: The notation [tex]H^r_{s} [/tex]is the number of weak compositions of **s** into **r** parts which is equivalent to the number of ways to select **s** objects from **r **types of objects with repetition allowed. ie. selecting **s** objects from the multiset [tex]\{\infty \cdot b_1, \ldots , \infty \cdot b_r \}[/tex]

Let

[tex]\displaystyle F(n) = \sum_{m=0}^N H^3_{2m} H^3_{n-2m} = \sum_{m=0}^N \binom{2m+2}{2}\binom{n-2m+2}{2}[/tex]

then [tex]F(1)= 3 , F(2) = 12, F(3) =28[/tex]. This means that the number of ways to get an even number from 1 die is 3 (agreed) and the number of ways to get an even sum from two dice is 12!

Now if you throw two dice, there are [tex]6 \times 6[/tex] possibilities, half of which is even, which means there should be 18 ways to get an even sum. The discrepancy lies in whether you consider each throw as distinct. I do and hence 3+1 and 1+3 give two ways to get a sum of 4. Thus my solution for the problem is: For **n** dice, we need to choose **2m** of them that must give odd numbers (3 choices each) and **n-2m** that must give even numbers (also 3 choices.) So

[tex] \displaystyle G(n) = \sum_{m=0}^N \binom{n}{2m} 3^{2m} 3^{n-2m} = \sum_{m=0}^N \binom{n}{2m} 3^{n} [/tex]

with [tex]G(1)=3, G(2)=18, G(3)=108[/tex].

To get the [tex]F(n) [/tex]given above, we would need to think in terms of multisets. The problem is choosing **n** numbers from 1 to 6 with repetition, such that total number of 1, 3 and 5 are even. Here the emphasis is that the sequence of choice is irrelevant. So we choose **2m** from [tex]\{\infty \cdot 1, \infty \cdot 3 , \infty \cdot 5 \} [/tex]and the remaining **n-2m** from

[tex] \{\infty \cdot 2, \infty \cdot 4 , \infty \cdot 6 \},[/tex] so

[tex] \displaystyle F(n) = \sum_{m=0}^N \binom{2m+2}{2}\binom{n-2m+2}{2}[/tex].