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Two tough distribution cases: 6 distinct chocolates to 4 children with everyone getting one (partitions 1-1-1-3 and 1-1-2-2) totals 1,560 ways; to 3 children (2-2-2, 4-1-1, 3-2-1) totals 540. Method: list valid compositions, assign recipients, then choose chocolates using combinations; sum across all cases carefully.
This article, along with its accompanying video, presents two high-difficulty examples of distributive arrangement problems, a concept that frequently appears in combinatorics. These examples focus on dividing objects among groups under strict restrictions, showing how small changes in conditions can dramatically alter the counting process. The aim is not to overwhelm with formulas but to highlight the logic and structure that allow such problems to be solved step by step. By analyzing cases like ensuring every participant receives at least one item, you will gain a deeper understanding of the variations and twists possible. Mastery of such advanced questions is an essential part of any serious GMAT preparation course, as it sharpens precision, adaptability, and clarity of reasoning.
Distribution problems become most challenging when every participant must receive something. For a detailed understanding, let us take two high difficulty questions
Let us carefully examine the case of dividing 6 different chocolates among 4 children, with the condition that each child gets at least 1 chocolate.
Here, three children get 1 chocolate each, and one child gets 3.
The child who receives 3 can be chosen in 4 ways.
Then, the chocolates are distributed: 6C1 × 5C1 × 4C1 × 3C3 = 120.
Multiplying, we get 4 × 120 = 480.
Here, two children receive 1 chocolate each, and two receive 2 each.
The children can be chosen in 6 ways (4! ÷ 2!2!).
Then, chocolates are distributed: 6C1 × 5C1 × 4C2 = 180.
Multiplying, we get 6 × 180 = 1080.
Total for these two scenarios: 480 + 1080 = 1560.
Therefore, the answer to this question is 1560.
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The possible variations are 2–2–2, 4–1–1, and 3–2–1.
Ways to choose children: 3!/3! = 1
Wats to distribute chocolates: 6C2 × 4C2 × 2C2 = 90.
Multiplying, we get 1 × 90 = 90.
Ways to choose children: 3!/2! = 3
Wats to distribute chocolates: 6C4 × 2C1 × 1C1 = 30.
Multiplying, we get 3 × 30 = 90.
Ways to choose children: 3! = 6
Wats to distribute chocolates: 6C3 × 3C2 × 1C1 = 60.
Multiplying, we get 6 × 60 = 360.
Combined total: 90 + 90 + 360 = 540.
Therefore, the answer to this question is 540.
The real value of these two distribution problems lies not in the numbers, but in the way they stretch your thinking. They reveal how a small change in conditions reshapes the entire approach, reminding us that formulas alone are never enough. You must learn to pause, to break cases logically, and to remain open to multiple paths of reasoning. These problems cultivate patience, flexibility, and attention to detail — all qualities that extend beyond test day. Regular exposure through drill and high number of GMAT mocks will help you internalize this mindset and face unpredictable challenges with clarity.
Challenging distribution problems remind us that complexity is often born from simple rules applied under constraints. In GMAT preparation, much like in these problems, the true skill lies in adapting to shifting conditions with calm logic. The same lesson applies in the application process for a competitive B-school such as ISB, where your strengths must be distributed carefully across essays, recommendations, and interviews. Life, too, is an exercise in distribution, where time, energy, and priorities must be balanced thoughtfully. Success comes not from abundance alone, but from how wisely we allocate what we already possess.
