Expanding my question on LS#17 M3

My question was answered about if x was left to y and y was left to z, but what if we expanded it further? (I'm going to refer the letters as ABCDEF to make it easier) If B has to be to the left of C, C to the left of D, D to the left of E, how much possibilities can be found out? Please answer with the number of possibilities and the fraction of the total. (I'm doing this to find a pattern)

We can still put permutations into groups. For example, with a permutation like BCAFDE, there are 23 other permutations in the group, each one switching the order of B, C, D, and E, while leaving A and F in the same spots. When we apply the same concept to other permutations, we can split all 6! total permutations into groups of 24. In each group, only one permutation satisfies the requirement that B is left of C, which is left of D, which is left of E: it's the one where the four letters are ordered B, C, D, E. And so in each group, 1 permutation works out of 24. So the number of permutations that work in total is 1/24 x 6! = 30 (I think).

@knowledgeabledog Thanks!