27. 8 letters, $A, B, C, D, E, F, G, H$ in all permutations, find the number of permutations where only 4 elements are not in their original positions.
27. 8 elements taken 4 at a time, there are $\mathrm{C}_{8}^{4}$ ways to choose, 4 elements staying in their original positions have only one way to be chosen, the other 4 elements not being in their original positions have $D_{4}$ ways, in total there are $\mathrm{C}_{8}^{4} \cdot D_{4}$ ways.