A dicey past year exam question

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]

[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].

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