SOLUTION: If 8 persons are seated at a random in a row. What is the probability that a certain couple (a) will be together (b) will not be together

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Question 1157242: If 8 persons are seated at a random in a row. What is the probability that a certain couple
(a) will be together
(b) will not be together

Answer by ikleyn(52909)   (Show Source): You can put this solution on YOUR website!
.

In all, there are 8! = 8*7*6*5*4*3*2 = 40320 permutations of 8 objects.


If two of them are together, we consider them as one item, and then we have, actually, 7! = 7*6*5*4*3*2 = 5040 different 
permutations for seven items, and two permutations inside this par; 

so, in all, there are 2*7! permutations for these 8 persons, where these two are together.


Therefore, the answers are


(a)  P =  =  =  =  = 0.25;


(b)  P =  1 -  =  = 0.75.

Solved.

------------------

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.


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