Combinatorics on "even outcomes"

Question

I read this problem:

Fifty identical (six sided) dice are rolled. How many distinct outcomes with even numbers of $1$'s, $2$'s, $\ldots$ , $6$'s are possible? (For example: an outcome might be eight $1$'s, fourteen $2$'s, ten $3$'s, ten $4$'2, two $5$'s and six $6$'s).

This problem really triggered me, yet I am not that good in combinatorics. If it were to calculate just all the possible outcomes, it would be an easy calculation. But how to take into account the even request? Each number must occur in an even amount.

Answer 1

For $i\in\{1,\dots,6\}$, let $x_i$ be the number of times that value $i$ appears. You want to count solutions to \begin{align} \sum_{i=1}^6 x_i &= 50 \\ x_i &\in \{0,2,\dots, 50\} \end{align} Perform a change of variables $y_i=x_i/2$, yielding \begin{align} \sum_{i=1}^6 y_i &= 25 \\ y_i &\in \{0,1,\dots, 25\} \end{align} Now perform the "easy" calculation.

Answer 2

The other answer is correct assuming the outcomes you are counting are entirely determined by the number of occurrences of each face, which seems to be the correct interpretation in this case. However, another reasonable interpretation would be to consider different sequences of the same number of outcomes to be distinct, so that with four dice, $(2,2,4,4)$ is different from $(4,2,4,2)$. This type of counting would be needed if you wanted to compute the probability of the event "every face occurs an even number of times" by dividing the number of outcomes by $6^{50}$.

In this interpretation, there is a nice, interesting solution: $$ \#\text{(outcomes where each face appears an even # of times)}=\boxed{\frac{6^{50}+6\cdot 4^{50}+15\cdot 2^{50}}{32}.} $$

To motivate the solution, let us consider an easier problem of counting the number of ways to flip $50$ coins and get an even number of heads, again with the order of the sequence of flips mattering. You can show that $$ (1+x)^{50}=\sum_{k=0}^{50}\#(\text{outcomes with $k$ heads})x^k. $$ Therefore, plugging in $x=1$ gives the total number of outcomes to be $(1+1)^{50}=2^{50}$, obviously. But we only want to count outcomes with an even number of heads, so how do we get rid of the summands counting sequences with an odd number of heads? A common trick is $$ \frac{(1+x)^{50}+(1-x)^{50}}{2}=\sum_{k=0}^{25}\#(\text{outcomes with $2k$ heads})x^{2k}. $$ Noe that averaging the original function of $x$ with the substitution of $-x$ cancels all odd powers of $x$ and preserves the even ones. Therefore, plugging in $x=1$ to this modified expression gives $$ \frac{2^{50}+0^{50}}2=2^{49}=\#(\text{outcomes with any even number of heads)} $$ To apply this to the die case, we instead have to use $$ (1+x+y+z+w+t)^{50}=\sum \#\text{(outcomes with $i$ ones, $j$ twos, $\dots$, $l$ fives)}x^i y^j z^k w^h t^l $$ We then have to average out all five of the variables to cancel out the summands where that variable appears an odd number of times, and finally plug in $1$ for each. The result is a sum of $32$ terms of the form $(1\pm x\pm y\pm z\pm w\pm t)^{50}$ all divide by $2^5=32$. When you substitute $1$ for each variable, the result is $$ \#\text{(outcomes where each face appears an even # of times)}=\frac{6^{50}+5\cdot 4^{50}+10\cdot 2^{50}+10\cdot 0^{50}+5\cdot (-2)^{50}+(-4)^{50}}{32} $$ which simplifies to the advertised answer.