.
(a) 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800 ways.
Any of 10 questions can be placed first;
any of remaining 9 questions can be placed 2-nd;
any of remaining 8 questions can be placed 3-rd
. . . . and so on . . . .
(b) We can consider two special questions as one single object in 2 states, that are its internal permutations AB and BA.
Then we have 9 objects and 9! their permutations with one the special object which has 2 states.
Therefore, in all, there are 2*9! different ways = 725760 ways.
(c) From the entire set of all possible permutations of 10 questions, subtract the number of (b) to get the answer
10! - 2*9! = 725760 = 3628800 - 725760 = 2903040.
Solved. All questions are answered.
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This problem is about PERMUTATIONS.
On Permutations, see introductory lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Problems on Permutations
- 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.