Description for Correct answer:
Number of arrangements = \( \large\frac{n!}{p! q! r!} \)

Total letters = 6, but R has come twice

So, required number of arrangements

= \( \large\frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 360\)