So I know the answer is
$$N(n) = \frac{( n + 5 )!}{ n! \cdot 5! }$$
where $n$ is the number of dice.
but I'm not able to prove it either mathematically or intuitively, although I'm able to verify for various cases.
I tried to prove it by first assuming the dice are different, and then eliminating duplicates but that method doesn't look very promising.
How do I prove the above result.
On
Let $n_1,\cdots,n_6$ be respectively the numbers of dice showing up $1,\dots,6$. We are looking for the number of non-negative solutions to the equation : $$n_1+\cdots+n_6=n.$$ Using stars and bars it is: $$ \binom {n+6-1}{6-1}. $$
Each outcome can be represented as a tuple of $6$ nonnegative numbers indicating how many times each side appeared (the numbers must sum to $n$). For instance, $(1,3,2,0,1,0)$ represents a roll of seven dice where "1" appears once, "2" appears three times, "3" appears two times, "4" appears zero times, "5" appears once, and "6" appears zero times.
You can count this using the stars-and-bars method: there are $\binom{n+5}{5}$ ways to arrange $5$ bars and $n$ stars in a row.