# Combinatorial Problem

Nathew
If we have the letters A, B, C, D, E, and F, and we are asked to find the number of arrangements where A is before B, wouldn't this just be half of the total number of arrangements with no restrictions? Intuitively this makes sense, but I have some doubts. For example, when I try to do the problem out, I get 2(5!)+2(4!)(2!)+(3!)(3!) which is slightly over half of the total arrangements with no restrictions.

Help me out please!

Simon Bridge