Animated Solution for Mathematics - Permutations and Combinations: 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 Combinatorial Journey

Welcome, future engineers. Today, we are not just solving a problem; we are embarking on a journey into the elegant world of combinatorics. We are tasked with a scenario that often trips up even the brightest students: distributing 5 distinct employees into 4 indistinguishable offices.

It sounds simple, but the moment you hear the word 'indistinguishable,' your mental alarm bells should ring. This is where the magic—and the danger—of counting begins.

Imagine you are standing in a hallway with five employees, let's call them $E_1,E_2,E_3,E_4,$ and $E_5$. You have four offices. If the offices were labeled 'Manager,' 'HR,' 'IT,' and 'Sales,' the problem would be straightforward.

But here, the offices are identical. If you put $E_1$ in one office and the rest in another, it doesn't matter which office $E_1$ is in. The only thing that matters is the grouping.

This shifts our perspective entirely. We are no longer looking for functions from a set of employees to a set of offices; we are looking for partitions. We need to break the number 5 into at most 4 parts, where each part represents the number of employees in a specific office.

To solve this, we must be systematic. We need to find all integer partitions of 5 that have at most 4 parts. Let's list them out:

1. $(5)$ 2. $(4,1)$ 3. $(3,2)$ 4. $(3,1,1)$ 5. $(2,2,1)$ 6. $(2,1,1,1)$

Each of these represents a unique way to group our employees. Now, we must calculate the number of ways to form these groups for each case.

Case 1: The Solo Act $(5)$ If all 5 employees sit in one office, there is only one way to do this. Since the offices are indistinguishable, it doesn't matter which office they choose. The number of ways is simply $\left(\genfrac{}{}{0ex}{}{5}{5}\right)=1$.

Case 2: The $(4,1)$ Split Here, we have one group of 4 and one group of 1. We choose 4 employees out of 5 to form the first group: $\left(\genfrac{}{}{0ex}{}{5}{4}\right)=5$. The remaining employee automatically forms the second group. Since the groups are of different sizes, there is no overcounting. We have 5 ways.

Case 3: The $(3,2)$ Split Similar to the previous case, we choose 3 employees out of 5 to form the first group: $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. The remaining 2 form the second group. Again, the sizes are distinct, so we have 10 ways.

Case 4: The $(3,1,1)$ Split We have one group of 3 and two groups of 1. We choose 3 employees for the first group: $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. From the remaining 2, we choose 1 for the next group: $\left(\genfrac{}{}{0ex}{}{2}{1}\right)=2$.

Because we have two groups of size 1, we must divide by $2!$ to account for the indistinguishable nature of these offices.

Case 5: The $(2,2,1)$ Split We have two groups of size 2 and one group of size 1. We choose 2 employees for the first group: $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. From the remaining 3, we choose 2 for the second group: $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$.

Again, we have two groups of size 2, so we must divide by $2!$ to account for the indistinguishable nature of these groups.

Case 6: The $(2,1,1,1)$ Split Finally, we have one group of 2 and three groups of 1. We choose 2 employees for the first group: $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. The remaining 3 employees each go into their own office.

Since we have three groups of size 1, we must divide by $3!$ to correct for the overcounting.

We have navigated the labyrinth of partitions. Now, we simply sum the results of our cases to find the total number of ways, $n$:

There you have it! By breaking the problem down into manageable, logical partitions and respecting the symmetry of the indistinguishable offices, we arrived at 51. Remember, in combinatorics, the most important step is often not the calculation itself, but the careful identification of the cases.