There are $16$ people separated into $4$ groups of equal sizes (each group has $4$ people).
In how many ways can we rearrange these people if two different people can't be in the same group twice (on different assignations)?
The answer is $4^3 \cdot 3^3 \cdot 2^3$, but I don't know how to prove it.
As we are given that those $16$ people are seperated initially in a particular arragement in $4$ equal groups of $4$ members each. *Condition-We need to rearrange them in such a way that no two persons can belong to the same group again as in the initial arrangement. Now we need to form our first group by choosing $4$ persons each from a different group of our initial arrangement. This can be done in $4^4$ ways.( By using fundamental principle of multiplication ) Now for forming the second group we are left with $3$ people in each group in our initial arrangement and the second group can be formed in $3^4$ ways. Similarly the $3$rd and $4$th group can be formed in $2^4$ and $1^4$ ways respectively. So the total number of ways=$\frac{4^4×3^4×2^4}{4!} We need to divide by 4! since the arrangement among the groups is not required here.
