Question 1208353
.
how many ways can 3 male students and 2 female students be arranged in 6 chairs 
around a round table in the following cases:
1) The female students are adjacent.
2) The female students are adjacent and the male students are adjacent.
3) No two female students are adjacent.
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        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
        basics of circular permutations.


        I will solve the problem in this order:  (1),  (3),  (2).



<pre>
(1) For part (1), we consider 2 female students as one block (= one glued object).

    I will call this object F.

    So, we have one free (unoccupied) chair, 3 male students and the block F: they are our objects
    to place them in some order around the table.

    Having round table and circular permutations around it, we can assume that the empty chair is 
    in position "North", or at "12 o'clock".  

    Then for other 4 objects (3 male students and block F) we have 4! = 4*3*2*1 = 24 distinguishable permutations.

    In addition, we have 2 (two) independent permutations inside the group of two blocked females.


    In all, it gives 2*24 = 48 different possible distinguishable circular permutations.    <U>ANSWER</U>


    At this point, part (1) is complete.




(3) For part (3), we have 3 + 2 + 1 = 6 objects (3 male students, 2 female students, and one free chair).

    For 6 objects around a table, there are (6-1)! = 5! = 120 different distinguishable circular permutations.

    From this set of permutations, we should subtract 48 permutations of part (1), where two female
    are adjacent.

    Doing it, we get the <U>ANSWER</U> for part (3): it is 120 - 48 = 72.




(2) For part (2), we have three objects: the block of 3 male; the block of 2 female, and the empty chair.

    For three objects, there are 3-1 = 2 (two) circular permutations.


    In addition to it, there are 3! = 6 independent permutations inside the block of 3 males 
    and 2! = 2 independent permutations inside the block of 2 females.


    In all, it gives  2*2*6 = 24 different distinguishable circular permutations.    <U>ANSWER</U>
</pre>

Thus all problems are solved and complete explanations are provided.