Distinguishable objects and indistinguishable boxes :
Such Questions are solved manually.. Don't use Formulas in this case as Method gives more intuition than just applying formula and getting answer.
Solving this type of Questions is similar to finding Partitions of a given set (in equivalence relations)
So, Go step by step so that you don't count same thing Twice or You don't miss anything.
Let employees be $1,2,3,4$
Since each office can contain any number of Employees...
1. Put All employees in One Office. (Similar to Partition $\left \{ 1,2,3,4 \right \}$) .. $1$ Way.
2. Put $Three$ emp in One office and $One$ employee in Different Office then $4$ Ways possible Just like $4$ Partitions Possible in case of Set Partition. .. Hence, $4$ Ways.
3. Put $Two$ emp in One office and $Two$ emp in Other... Hence, $\binom{4}{2} / 2 = 3$ Ways. Just like Set Partition in Finding Number of Equivalence relations.
4. Put $Two$ Emp in One office, $One$ emp in Second, And $One$ emp in Third office... Hence, $\binom{4}{2} = 6$ Ways.
Next Case that could be possible in Set Partition is Put All employees(elements) in Different-Different Offices(Sets).. But that is Not possible here as there are Only $3$ Offices.
Hence, Total Ways Possible = $1 + 4 + 3 + 6 = 14$ Ways
Just Practice this approach for "Distinguishable objects and indistinguishable boxes" scenario and You will never do any question wrong.
Result : If there are $m$ Distinguishable objects and $n$ indistinguishable boxes and $n \geq m$ then Number of ways to put $m$ objects into $n$ indistinguishable boxes,when each box can contain any number of objects will be Same as Number of Partition of a Set with $m$ elements.
