in how many no.of ways do 4 men and 2 women seated in circular table that women not allowed to sit together ?

Question

permutation sum plz help

Answer 1

6 unless you are alowing people on the table. how much for a dance?

Answer 2

Assuming that you don't count different seating arrangements if you rotate the table then I get 4!*3 = 72

Start with one fixed location. I start with a women. The reason I do that is because you can always rotate the table until a specific person is in a designated location. The other women can only be in one of three locations

w1, m, w2, m, m, m
w1, m, m, w2, m, m
w1, m, m, m, w2, m

Then calculate the ways that the 4 men can be seated in the 4 open locations. Which ends up being 4 factorial.

If you do count different seating arrangements when you rotate then you multiply that by 6 to get 432.

Answer 3

Imagine the table with the positions :

1 - 2 - 3 - 4 - 5 - 6

where 6 is near of 1.

Clearly your answer depends if you consider or not the next ways of sitting people equal or not (this is a particular case):

M1 - M2 - W1 - M3 - M4 - W2

W2 - M1 - M2 - W1 - M3 - M4

So the question is , does it matters in which chair is the people seated or not ?

The solution is different. Add details.

Answer 4

There is information left out of your problem! Are men and women each fungible? (Can one man stand for another or are they unique?) Are rotations of the positions on the table considered equivalent? Is flipping the table position across an axis of symmetry equivalent?

Considering all of those equivalences, then the problem is quite simple. Take all of the possibilities, including the one where the two women are sitting in adjacent seats.

w1 w2 m1 m2 m3 m4
w1 m1 w2 m2 m3 m4
w1 m1 m2 w2 m3 m4

All other possibilities are equivalent to these three. We can eliminate the first case, since two women are juxtaposed, which leaves only two possibilities.

If the rotation and inversion equivalence classes that I have defined do not hold for your problem, then just generate the the possibilities from each of these three cases, by rotating each 5 times (the 6th rotation will return us to the original case), and then flipping the circular diagram over and rotating that 5 times. Doing this, you will discover that some possibilities are the same as others (since we are still using interchangeable men and interchangeable women). There are in fact only 6 distinct possibilities with women adjacent, 6 with one man in between, and 3 with two women in between. (I show why there are only 3 in the next paragraph.) So that makes 9 possibilities.

If individual men (or women) can change places to create a new permutation, then the number of possibilities increases greatly! For example, w1 m1 w2 m2 m3 m4 would be counted separately from w2 m1 w1 m2 m3 m4. In this case, let's back up to our original three possibilities. Let's say that a woman walks in the room first. She has 6 chairs to choose from. Then woman number two enters. In the case where she must sit adjacent to woman number one, then she has only two chairs to choose from. She can either sit to the first woman's right or her left. (This is also true of the case where she must sit with one man between herself and woman number one.) Then the men enter, first with 4 chairs to choose from, next with 3, next with 2, and finally only one chair remains for the last guy. So for the first two of our three seatings listed above, each has 6 * 2 * 4 * 3 * 2 * 1 = 288 permutations. The last case is different, however. When woman number two enters, she must sit across from woman number one. She thus has no choice about what chair to sit in. There is therefore half the previous number of permutations in this third solution to our seating dilemma: 6 * 1 * 4 * 3 * 2 * 1 = 144. (Notice that 288 + 288 + 144 = 720 = 6! = all possible permutations of six people in six chairs.) Hence there are 288 + 144 = 432 possible permutations with women apart, and 288 permutations with women together.

What is more interesting is the question of how you distribute m women as evenly as possible around a table with n men. This is trivial in the present case of 2 women and 4 men. The solution is w1 m1 m2 w2 m3 m4, with the women sitting opposite one another, but when m and n are relatively prime, the problem becomes much more interesting! The general solution is surprisingly complicated mathematically, but it is intuitively quite simple to figure out the solution for each case. For the solution, see Clough, John and Douthett, Jack (1991). "Maximally Even Sets", Journal of Music Theory 35: 93-173.

Answer 5

the formula for circular permutation is (n-1)! if the two women cannot sit together, (6-1-1)! = 24


maybe

Answer 6

8*4!