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Number of arrangements n ! /( p ! q ! r ! )Total letters = 6, but R has come twice.So, required number of arrangements = 6 ! / 2 ! = (6 x 5 x 4 x 3 x 2 !) / 2 ! = 360
Description for Correct answer: Number of arrangements = \( \large\frac{n!}{p! q! r!} \)Total letters = 6, but R has come twiceSo, required number of arrangements = \( \large\frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 360\) Part of solved Permutation and combination questions and answers : >> Aptitude >> Permutation and combination Comments Similar Questions
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