Suppose you are given the digits 1, 1, 2, 2, 3, 3, 4, 4. How would you find the number of 3-digit permutations you can create from those?
For 8-digit permutations the answer should be $\frac{8!}{2!2!2!2!}$ I believe.
However, I cannot think of a way to do this when asked for permutations where not all of the set is used. I have tried $\frac{8P3}{2!2!2!2!}$ but that does not match the amount when I try manually listing all permutations which seems to be 60.
If the problem were to be generalized what would a solution look like?
Thanks in advance.