I am currently doing this two part question and I am confused about the second part, can someone help explain how to do it?
Part a)
How many derangements of $$1,2,3,4,5,6,7,8 $$ start with 1,2,3, and 4 in some order?
My answer: I believe it is 81 because 1-4 have 9 derangements according to the Inclusion Exclusion formula and 5-8 also have 9 derangements so it's simply 9*9 = 81.
Part b)
How many derangements with 5,6,7, and 8 in some order?
My answer: I wrote it was 81 as well, but apparently I am wrong, can someone explain to me why the answer is $$(4!)^2$$
I just can't seem to understand what the difference is.
Best Answer
I assume you mean "how many derrangements starting with 5, 6, 7, and 8".
Every sequence that starts with those 4 values is a derangement. So all you are doing is counting how many sequences there are total, which is 4! for the first half and 4! for the second half.