Grouping And Division
The number of ways to divide 12 distinct books into three groups containing 4, 4 and 4 books respectively, where the groups are unlabelled and indistinguishable, equals which value?
Select the correct option:
Solution
5775
Dividing distinct objects into equal unlabelled groups requires dividing the labelled count by the factorial of the number of equal groups, a subtle JEE Advanced correction. The number of ways to split 12 books into ordered groups of 4, 4 and 4 is the multinomial 12!/(4!·4!·4!) = 34650. Since the three groups are of equal size and unlabelled, each distinct division has been counted 3! = 6 times by permuting the identical-sized groups. Dividing by 6 gives 34650/6 = 5775. Option 34650 is the labelled count, not the unlabelled one. Option 1925 divides by an incorrect factor. Option 11550 divides by only 3 instead of 3!. Hence there are 5775 ways. Plausibility check: the correction factor must equal the number of ways to permute the equal groups, which is 3! = 6, and 34650/6 = 5775 confirms that swapping identical-sized groups does not create new divisions. Stars and bars is the definitive tool for distributing identical objects into distinct groups, and the at-least-one condition is dispatched by reserving the minimum allotment before applying the free-distribution count. The method counts non-negative integer solutions of a linear equation by arranging dividers among identical items, a translation that turns many word problems into a single binomial coefficient.
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- grouping and division
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
5775
Dividing distinct objects into equal unlabelled groups requires dividing the labelled count by the factorial of the number of equal groups, a subtle JEE Advanced correction. The number of ways to split 12 books into ordered groups of 4, 4 and 4 is the multinomial 12!/(4!·4!·4!) = 34650. Since the three groups are of equal size and unlabelled, each distinct division has been counted 3! = 6 times by permuting the identical-sized groups. Dividing by 6 gives 34650/6 = 5775. Option 34650 is the labelled count, not the unlabelled one. Option 1925 divides by an incorrect factor. Option 11550 divides by only 3 instead of 3!. Hence there are 5775 ways. Plausibility check: the correction factor must equal the number of ways to permute the equal groups, which is 3! = 6, and 34650/6 = 5775 confirms that swapping identical-sized groups does not create new divisions. Stars and bars is the definitive tool for distributing identical objects into distinct groups, and the at-least-one condition is dispatched by reserving the minimum allotment before applying the free-distribution count. The method counts non-negative integer solutions of a linear equation by arranging dividers among identical items, a translation that turns many word problems into a single binomial coefficient.
This hard difficulty mathematics question is from the chapter permutations and combinations, covering the topic of grouping and division. It appeared in the 2025 exam.
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