From 5 violinists, 5 violists, 5 cellists, and 5 pianists, how many ways can 5 quartets be formed?
Let

the violinists be $\quad a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$

the violists be $\quad b_{1}, b_{2}, b_{3}, b_{4}, b_{5}$

the cellists be $\quad c_{1}, c_{2}, c_{3}, c_{4}, c_{5}$

the pianists be $\quad d_{1}, d_{2}, d_{3}, d_{4}, d_{5}$

Grouping those in the same column, we get 5 quartets. Each row's elements can be written in 5! different orders, and by connecting the elements of the resulting permutations column-wise, we can obtain 5 quartets in $(5!)^{4}$ ways, always keeping the elements of each column together.

Even now, if we consider certain 5 columns, we get the same 5 quartet arrangements by keeping the elements in the column and arranging them in 5! different orders. Thus, the number of possible arrangements is: $\frac{(5!)^{4}}{5!}=(5!)^{3}$.