JEE Mathematics 2024 Question with Solution

Given: $5$ different employees and $4$ indistinguishable offices, with empty offices allowed.

Find: The number of ways, $n$, to distribute the employees.

Since the offices are indistinguishable, we count partitions of $5$ distinct employees into at most $4$ non-empty unlabeled groups.

Using Stirling numbers of the second kind:

Therefore,

Alternatively, from the partition listing shown:

• $5,0,0,0$ gives $1$ way
• $4,1,0,0$ gives $\frac{5!}{4!} = 5$ ways
• $3,2,0,0$ gives $\frac{5!}{3!2!} = 10$ ways
• $2,2,1,0$ gives $\frac{5!}{2!2!1!} = 15$ ways
• $2,1,1,1$ gives $\frac{5!}{2!1!1!1!} = 10$ ways
• $3,1,1,0$ gives $\frac{5!}{3!1!1!} = 10$ ways

Adding these,

Therefore, the correct option is C.

Given: $5$ distinct employees and $4$ indistinguishable offices.

Find: The value of $n$.

The total number of partitions of $5$ distinct elements is the Bell number $B_5$.

From the solution,

The case $S(5,5)=1$ corresponds to all $5$ employees sitting separately, which would require $5$ non-empty offices. That is not possible because only $4$ offices are available.

So,

Therefore, the number of ways is $51$, so the correct option is C.