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(Dividing by repetitions)
If you have n items where n₁ are of type 1, n₂ are of type 2, ..., nₖ are of type k, then the count is:
n! / (n₁! × n₂! × ... × nₖ!)
Example: AABBCC has 6 letters (2 A's, 2 B's, 2 C's). Count = 6! / (2! × 2! × 2!) = 720 / 8 = 90.
The denominator cancels duplicate arrangements.
