.
(a) Considering five couples as five special objects,
we have 5! = 5*4*3*2*1 = 120 permutations of these objects.
Next, we have 2 possible permutations (A,B) --> (B,A) inside each of 5 pairs.
These permutations are independent --- so, there are = 32*120 = 3840 permutations of 5 couples, satisfying the imposed conditions.
(b) If no restrictions are imposed, then there are 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800 possible permutations of 10 persons.
Solved.
----------------
On Permutations, see introductory 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.