<s>[INST] In how many ways can $m$ employees be assigned to $n$ projects if every project is assigned to at least one employee? In how many ways can we assign $m$ employees to $n$ projects so that every employee is assigned to exactly one project and every project is assigned to at least one employee? Also, $m>n$.
My answer:
Let's choose $n$ employees from $m$ and assign them projects such that each project has exactly one employee working under it. Thus, $mPn$.  For the remaining $m-n$ employees, each employee can work under $n$ project options. Thus, $n^{m-n}$.  Thus, total answer $= mPn \cdot n^{m-n}$.
For $m = 5$ and $n = 4$, my answer $= 480$ and expected answer $= 240$.
For $m = 6$ and $n = 3$, my answer $= 3240$ and expected answer $= 540$.
 [/INST] For $m=5$, $n=4$ . . .
Exactly one project gets two people, the others get one.


*

Choose the project with $2$ people: $\binom{4}{1}=4$ choices.

Choose the $2$ people for that project: $\binom{5}{2}=10$ choices.

Assign the remaining $3$ people to the remaining $3$ projects: $3!=6$ choices.

Hence the number of valid assignments is $(4)(10)(6)=240$.

For $m=6$, $n=3$ . . .

Up to reordering, the number of people for the $3$ projects is one of
\begin{align*}
1,1,4\\
1,2,3\\
2,2,2\\
\end{align*}
Consider each case separately . . .

Case: $1,1,4$.


*

Choose the project with $4$ people: $\binom{3}{1}=3$ choices.

Choose the $4$ people for that project: $\binom{6}{4}=15$ choices.

Assign the remaining $2$ people to the remaining $2$ projects: $2!=2$ choices.

Hence the number of valid assignments is $(3)(15)(2)=90$.

Case: $1,2,3$.


*

Choose which project gets how many people: $3!=6$ choices.

Choose the $3$ people for the $3$-person project: $\binom{6}{3}=20$ choices.

Choose the $2$ people for the $2$-person project: $\binom{3}{2}=3$ choices.

Hence the number of valid assignments is $(6)(20)(3)=360$.

Case: $2,2,2$.


*

Choose the $2$ people for the first project: $\binom{6}{2}=15$ choices.

Choose the $2$ people for the second project: $\binom{4}{2}=6$ choices.

Hence the number of valid assignments is $(15)(6)=90$.

Thus, the total number of valid assignments is $90+360+90=540$.
</s>