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1. In how many ways can 3 male students and 2 female students be seated on 6 chairs
arranged in a circle such that no two male students are adjacent?
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In this my post, I will solve problem 1, only.
As this problem is posed (with one vacant chair), I may assume that
you are just familiar with more simple problems without vacant chair
and know the basics of placement around circular table, as well as
the basics of circular permutations.
We have 6 chairs, in all. Of them, one chair is vacant, 3 male students sit on their chairs
such that no two male students are adjacent, and two female students sit on their chairs.
Having round table and circular permutations around it, we can assume that the vacant chair is
in position "North", or at "12 o'clock" (position n1).
Then for 3 male students and two female students we have chairs in five remaining positions 2, 3, 4, 5, and 6.
Since no two the male students are adjacent, it means that three male students occupy 3 chairs with even numbers 2, 4, 6,
while two female students occupy 2 chairs with odd numbers 3 and 5.
Then for three male students we have 3! = 6 possible permutations on their chairs, and for two female students
we have 2! = 2 possible permutations on their chairs.
In all, it gives us all 6*2 = 12 possible distinguishable circular arrangements
satisfying the imposed restrictions.
Solved.