How many distinct rearrangements can be made from the letters of the word CARRIER?

1. CARRIER has 7 letters: C, A, R, R, I, E, R. 2. If all letters were distinct there would be 7! = 5,040 orderings. 3. The letter R repeats 3 times, and rearranging identical letters among themselves changes nothing. 4. Divide by the 3! = 6 internal orderings of the three Rs. 5. The count of distinct arrangements is 7!/3! = 5,040/6. 6. That gives 840 distinct rearrangements of the word. 7. Option B, 5,040, never corrects for the repeated Rs, so it counts each true arrangement six times. 8. Option C divides by 3 instead of 3!, and option D divides by 2! as if only two letters repeated. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.5 COUNTING PRINCIPLES_