Rolling $k$ identical dice, how many possibilities are there?

Question

I understand that the solution is $\binom{6-1+k}{k}$, the same as the solution to distributing $k$ balls into $6$ boxes, but my first thought was to solve it differently, and I can't understand why it's incorrect.

My thought was, to describe the result of throwing $k$ dice as a string of $k$ letters, each letter representing the result of one die. So we get $6^k$ different strings, but the order of the letters doesn't matter because the dice are identical, so we divide by $k!$.

Why is this incorrect?

Answer 1

So you should be careful about what you divided, when you said

$$\frac{6^k}{k!},$$

you were actually assuming that all $6^k$ possibilities have the same number of permutations all represent the same combination, i.e. $k!$, in common. But say there are six dices and one of those $6^6$ results is

$$123455,$$

then by your formula you would divide it by $6!$, but this means you should find all others $6!-1$ ways which all represent the same combinations of "one $1$, one $2$, one $3$, one $4$, two $5$", but in fact there are just

$$\frac{6!}{2!}-1$$

other cases.

Answer 2

When there are multiple copies of a number in the string it cannot be reordered in $k!$ ways. Think of flipping a coin three times. There are $2^3=8$ ordered results of the flips. You would claim there are $\frac {2^3}{3!}=\frac 43$ unordered results when in fact there are $4$-the number of heads can be $0$ to $3$. There is only one string that has $0$ heads, three that have $1$, and so on.