If you treat all the "e's" as different letters (i.e. consider the number of permutations of $e_1, b, e_2, n, e_3, z, e_4, r$ there are $8!$ different ways of arranging this. However there are $4!$ ways of arranging $e_1, e_2, e_3, e_4$, so these permutations occur in "groups" of $4!$ permutations which are in fact the same permutation. Hence the answer is $\frac{8!}{4!}$.