let ur mind work

abc

6 men with their wives ( total of 12 ) in how many ways can they sit in a circular table but no man sits beside his wife ????
needs smart people
cheers
abc

T@P

just a hint in case you are totally lost, a good approach would be to find the numebr of ways 12 people can sit around a table, and then subtract the numebr of times a man and his wife are nxt to each other.
then when a man and his wife ( 1 pair) sit together, every combination where tey dont move is bad. subtract the nume ways 10 people can sit around a table. then when two pairs of husband wives are together, so 8 people around a table etc. doing it out, you should come at the right answer. unfortunately, i never really came across any 1 step ways of doing these kinds of problems, but if you know one i would *very* interested in knowing it

Bartholomew

It's more complicated than that because when you subtract those arrangements with one particular pair (say man 1 and wife 1) together, you are also subtracting some arrangements with man 2 and wife 2 together, and you only want to count each arrangement once. I think, although I have not really thought about it, that you could do it the way I ended up doing the first of "Two Combinatorial Problems" in the set theory, logic, probability and statistics forum.

gazzo

Hmmm.... i dunno if its right.

Select if you want to see: "115200"



If each person represented a vertex of a regular 12-gon, could you solve it with err.. hmm.

Gokul43201

I think not. I believe the number will be larger than 500,000. Nearly all the ways of seating the six men together, and their wives, together on the other side, will produce a successful arrangement. This can be done in (6!)^2 ways, which is bigger than 500,000.

Moses

Well, a good tactic you may use is to make pair 1 man set next to par 2 women, and on the opposite side of the circle the opp will happen [pair 1 women, will sti next to pair 1 man].

after words let the men set on one side [count the !]. then the women on the other side , then do it as 3 men, 1 women, and so one. After words, instead of using the paris 1-2 as "walls" use 1 & 3 and so on, so the number is soo mad large. Well, i am sure with factorials you can make the calulating less paining, but still a pain.

mattmns

Question: If 6 men sit next to each other, and 6 women sit next to each other (at the same time) Is this counted as 1 possibility, or 12? Meaning that you would move each man and woman one seat, for example, left (11 times).

Gokul43201

It's 1 arrangement (not 12). We are not told that the seats are numbered, so must assume they are identical.

Rogerio

Just 1 possibility. Remember it is a circular table.

Galileo

But the men and women ARE numbered right? So permuting (or is it permutating?) the men will give a different arrangement and permut(at)ing the women will as well.
So there are (6!)^2 ways of arranging the table such that 6 men are sitting next to each other.

EDIT: Nevermind, I don't think there was any confusion...

mattmns

Ok I thought it was 1, but your number of 500,000 got me questioning that. Now back to the problem.

Rogerio

And it is! 12,771,840

hemmul

i got 12!-4*6!
if it is true - i'll explain the solution...

Rogerio

Unfortunately it's false: it can't be more than 11!

hemmul

whoops! so i misunderstood something...

b0mb0nika

look at the problem the other way around :
how many ways can the 6 men ( and women ) sit at the round table such that every men sits besides his wife :
there are 5! x 2^6 possible way to arrange the 6 pairs ( look at the pair - man + woman as 1 object, there are 5! ways you can arrange them around a round table, multiply it by 2^6 b/c for each pair it doesn't matter if the man or the woman is on the left/ right side)

now we know that 12 people can be arranged in 11! ways around a round table
so there are

11! -5! x 2^6
ways to arrange the people such that no man sits beside his wife

i believe that is the answer
Diana

Rogerio

Well Diana, if "no man sits beside his wife", then "only one man beside his wife" is not allowed.
However, this arrangement doesn't belong to your exceptions set.
So, your answer is not the correct one.

BTW, I think it is 12,771,840