Question 1194573
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First person can seat at any of 12 seats, giving 12 options.

Second person can seat at any of remaining 11 seats, giving 11 options.

Third person can seat at any of remaining 10 seats, giving 10 options.

Fourth person can seat at any of remaining 9 seats, giving 9 options.


The total number of possible seating arrangements is  12*11*10*9 = 11880.    <U>ANSWER</U>


It is the product of 4 consecutive integers numbers in descending order, starting from the number of 12.
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Solved and explained.


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To see many similar &nbsp;(and different) &nbsp;solved problems, &nbsp;look into the lesson


&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Special-type-permutations-problems.lesson>Special type permutations problems</A> 


in this site.  &nbsp;&nbsp;Learn the subject from there.