1. Permute 2 objects of 2 distinct types in 4 spots.
You're placing 2 A-type and 2 B-type objects into 4 positions. The number of distinct permutations is:
\[
\frac{4!}{2! \cdot 2!} = 6
\]
Memory Trick:
Think of the 4 spots as a 4-letter word made of 2 As and 2 Bs (e.g., AABB).
To remember the count:
Total ways to arrange 4 items = 4!
But A and B repeat, so divide by repeats:
* 2! for A's
* 2! for B's
* So: $$
\frac{4!}{2! \cdot 2!} = 6
$$