I have a question here which goes:
8 speakers are to speak one after another in a conference. Dr Cooper’s lecture is related to Dr Hofstadter’s and should not precede it. How many schedules could be arranged?
The solution is 8!/2 = 20160 which I understand how it is derived through the formula of permutation of non-distinct objects.
However, my first initial solution is 6! * 7P2 = 30240
Reasoning being, The 6 people that do not have any restriction I will arrange them first with order considered. Thereafter I slot the two other (Cooper and Hofstadter) in the 7 slots within the 6 people. I dont get which test case did I double counted in using this formula.
Please advice.
You used $7P2$, that means Dr C and Dr H can come in any order but Dr C's should come after Dr H's lecture. You should use $7c2$.
Also, you're missing cases like HC123456 and 652HC134