.
8 items have 8! = 8*7*6*5*4*3*2*1 = 40320 permutations.
But in this problem, we consider not all the permutations.
We consider only THOSE, where the most difficult question and easiest question are set next to each other.
So, these two questions always are neighbor.
It means, that we consider these two objects as two clued together;
so we actually consider 7 objects, and this special object can exist in two states (ED) or (DE), depending
which one of the two questions, easy (E) or difficult (D), goes first.
THEREFORE, the total number of all such possible permutations is
2*7! = 2 * (7*6*5*4*3*2*1) = 2 * 5040 = 10080. ANSWER
Solved.
==============
If you want to see many similar solved problems on PERMUTATIONS, see my lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Simple and simplest problems on permutations
- Special type permutations problems
- Problems on Permutations with restrictions
- OVERVIEW of lessons on Permutations and Combinations
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.