.
According to the condition, the two questions are FIXED at their permanent positions; they do not participate
in permutations.
Only 6 remaining questions do participate.
For them, there are 6! = 6*5*4*3*2*1 = 720 possible permutations.
ANSWER. Under given conditions, there are 6! = 720 possible permutations.
Solved.
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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.