Permutations with Indistinguishable Items Given a list of integers, output the number of permutations of the integers, with indistinguishable permutations counted once. If there are n integers, and each group of indistinguishable numbers has length n_i, this is n! / (n_1! * n_2! * ...)
Rules

Input will be some form of list as arguments to a function or the program with 1 to 12 non-negative integers.

Output will be printing or returning the number of permutations as described above.

No standard loopholes or built-in functions (generating permutations, combinations, etc.). Factorials are allowed.


Test Cases
Inputs:
1, 3000, 2, 2, 8
1, 1, 1
2, 4, 3, 2, 3, 4, 4, 4, 4, 4, 1, 1

Outputs:
60
1
83160
Python, 48 bytes
f=lambda l:l==[]or len(l)*f(l[1:])/l.count(l[0])

A recursive implementation.
In the formula, n! / (n_1! * n_2! * ...), if we remove the first element (say it's 1), the number of permutation for the remaining n-1 elements is
(n-1)! / ((n_1-1)! * n_2! * ...) ==
n! / n / (n_1! / n_1! * n_2! * ...) == 
n/n_1 * (n! / (n_1! * n_2! * ...)`)

So, we get the answer by multiplying by n/n1, the reciprocal-fraction of elements that equal the first one, by the recursive result for the rest of the list. The empty list gives the base case of 1.