Permutations With Repetition
The number of distinct arrangements that can be formed using all the letters of the word BANANA, accounting for repeated letters, equals which value?
Select the correct option:
Solution
60
When some objects are identical, the number of distinct permutations is the total factorial divided by the factorials of the repetition counts, a core JEE Advanced formula. The word BANANA has 6 letters: A appears 3 times, N appears 2 times, and B appears once. The number of distinct arrangements is 6! divided by (3! × 2! × 1!) = 720 / (6 × 2 × 1) = 720 / 12 = 60. Dividing by the factorials of repeated letters removes the overcounting caused by permuting identical letters among themselves. Option 720 = 6! treats all letters as distinct. Option 120 = 5! mishandles the repetition. Option 360 divides by only one repetition factor. Hence there are 60 distinct arrangements. Plausibility check: the count must be smaller than 6! = 720 by exactly the factor 3! × 2! = 12 that accounts for indistinguishable rearrangements, and 720/12 = 60 confirms the multinomial correction.
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- permutations with repetition
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
60
When some objects are identical, the number of distinct permutations is the total factorial divided by the factorials of the repetition counts, a core JEE Advanced formula. The word BANANA has 6 letters: A appears 3 times, N appears 2 times, and B appears once. The number of distinct arrangements is 6! divided by (3! × 2! × 1!) = 720 / (6 × 2 × 1) = 720 / 12 = 60. Dividing by the factorials of repeated letters removes the overcounting caused by permuting identical letters among themselves. Option 720 = 6! treats all letters as distinct. Option 120 = 5! mishandles the repetition. Option 360 divides by only one repetition factor. Hence there are 60 distinct arrangements. Plausibility check: the count must be smaller than 6! = 720 by exactly the factor 3! × 2! = 12 that accounts for indistinguishable rearrangements, and 720/12 = 60 confirms the multinomial correction.
This easy difficulty mathematics question is from the chapter permutations and combinations, covering the topic of permutations with repetition. It appeared in the 2025 exam.
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