JEE Main 2024

Question

If $n$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $n$ is equal to A). $47$ B). $53$ C). $51$ D). $43$

Edvaya · Accepted Answer

The correct answer is $51$. **Solution:** We need to distribute 5 distinct employees into 4 indistinguishable offices where any office can contain any number of employees (including zero). This problem can be solved by finding the number of ways to partition the integer 5 into at most 4 parts, and then calculating the number of ways the 5 distinct employees can be divided according to those partitions. The possible partitions of 5 into at most 4 non-empty parts are: 1. **[5, 0, 0, 0]**: All 5 employees in 1 office. Ways $= \frac{5!}{5!} = 1$ 2. **[4, 1, 0, 0]**: 4 employees in one office, 1 in another. Ways $= \frac{5!}{4! \cdot 1!} = 5$ 3. **[3, 2, 0, 0]**: 3 employees in one office, 2 in another. Ways $= \frac{5!}{3! \cdot 2!} = 10$ 4. **[3, 1, 1, 0]**: 3 employees in one office, and 1 each in two other offices. Ways $= \frac{5!}{3! \cdot 1! \cdot 1! \cdot 2!} = 10$ (Note: divide by 2! because the two groups of 1 are identical in size) 5. **[2, 2, 1, 0]**: 2 employees each in two offices, 1 in a third. Ways $= \frac{5!}{2! \cdot 2! \cdot 1! \cdot 2!} = 15$ 6. **[2, 1, 1, 1]**: 2 employees in one office, and 1 each in the remaining three offices. Ways $= \frac{5!}{2! \cdot 1! \cdot 1! \cdot 1! \cdot 3!} = 10$ Summing all the valid distributions: $n = 1 + 5 + 10 + 10 + 15 + 10 = 51$.

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