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.
Found 2 solutions by ikleyn, mccravyedwin:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~
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).
(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. ANSWER
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 ANSWER 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. ANSWER
Thus all problems are solved and complete explanations are provided.
Answer by mccravyedwin(407)  (Show Source):
You can put this solution on YOUR website!
We will first consider only 5 chairs and 5 people, and then think of how to
insert a vacant chair between two of the people appropriately.
[This is a round table problem. I always complain about round table problems.
That's because in every one of them, we must pretend the totally unrealistic
situation that the table, chairs, and people are all placed on a huge turntable
that can be rotated in any direction without considering any rotation of the
huge turntable to be a different seating arrangement. We know this is not the
way it is in reality, but we have to live with this anyway, because every math
book I've ever seen considers round table problems to be this unrealistic way.
So we'll assume it here.]
With that huge turntable in mind, there are only two possible configurations,
sex-wise, illustrated below, before the vacant chair is inserted. You may like
to contemplate whether there is any other possible configuration, but you'll
find that it will always be possible to rotate the huge turntable to show that
it's always and only one of these two.
 or 
1) The female students are adjacent.
It can only be like the configuration on the left above, but we may only
insert the vacant chair in any of the 4 places that are NOT between the
two females.
Arrange the females 2!=2 ways
Arrange the males 3!=6 ways
Insert the vacant chair any of 4 ways.
Answer: (2)(6)(4) = 48 ways.
--------------------------------------------------
2) The female students are adjacent and the male students are adjacent.It can also only be like the configuration on the left above, except this
time the vacant chair can only be inserted in 2 ways, between a male and a
female.
Arrange the females 2!=2 ways
Arrange the males 3!=6 ways
Insert the vacant chair either of 2 ways.
Answer: (2)(6)(2) = 24 ways.
--------------------------------------------------
3) No two female students are adjacent.
It can be like either of the configurations above, and thus there are two cases.
or
First case: It's like the configuration on the left, and the vacant chair can
only be inserted between the two females to keep them from being adjacent.
Arrange the females 2!=2 ways
Arrange the males 3!=6 ways
Insert the vacant chair only 1 way, between the two females
Answer: (2)(6)(1) = 12 ways.
Second case: It's like the configuration on the right, the vacant chair can
be placed between any two people, i.e., 5 places.
Arrange the females 2!=2 ways
Arrange the males 3!=6 ways
Insert the vacant chair any of 5 possible ways.
Answer: (2)(6)(5) = 60 ways.
Total for the two cases 12+60 = 72 ways.
Edwin
|
|
|
