Question

Good evening. I fail to comprehend, and would like your kind help:

A group of 12 people consisting of 6 married couples are to be seated together in a straight row.

1. How many different ways are there of sitting the 6 married couples if a man and his wife are seated next to each other?

2. How many different ways are there of seating the 12 people if no two women are seated together?

Answer 1

It is a question of permutations I think

which is I guess:

\[6P _{2} =\frac{ 6! }{ 2!} = 360 \]

there are about 360 ways !

I think the answer to 1 first is also answer to II 'cuz:

If 6 couples are seated in xyxyx..order[x= man y=woman)
and if no 2 woman are allowed to sit together than it is also same as:
xyxyxyxyx...

I'm not quite sure..what's the right answer in your book @kirinoarashi

Answer 2

The answer to 1. is 46080, while 2. is 3628800, unfortunately. Thank you for your attempt, tough.

Answer 3

there are 6! ways to seat the couples, and each seating has 111111(binary) states that it can exist as for example:

aa' bb' cc' dd' ee' ff' <--> 000000
aa' bb' cc' dd' ee' f'f <--> 000001
aa' bb' cc' dd' e'e ff' <--> 000010
aa' bb' cc' dd' e'e f'f <--> 000011

etc ....

Answer 4

and each couple may be seated in 2 ways: a a' or a' a

so 6! 2^6

Answer 5

yep :)

Answer 6

part 2 is more complicated

Answer 7

one could argue that it is just as complicated, but more may be fitting as well lol

Answer 8

if we seat them as before:

aa' bb' cc' dd' ee' ff'

there are 6! ways to seat the women, and 6! ways to seat the men

and then we can swap them as a'a b'b c'c ... in which there is 6! ways each to seat them again

Answer 9

not sure if im missing something on that one, or if the answer given is an error

Answer 10

at first I thought they were still being seated as couples (with no two women seated next to each other). But a better interpretation is all 12 are seated with no restrictions except no two women together.

Answer 11

For Part 2: how about 6! 6! 2

Answer 12

If we just use M and W just to see the pattern, and seat them like
MWMWMWMWMWMW (1) or flip them
WMWMWMWMWMWM (2)

But then for (1) we could consider:
WMMWMWMWMWMW (the left end end changed)

And for (2):
WMWMWMWMWMMW (the right end changed)

is there anything else that could be considered

Answer 13

each person is different, so there are lots of ways to arrange each of your patterns

Answer 14

oh I know each person is different, but I was just trying to see the ways in which 2 women wouldn't be seated together. I know my picture doesn;t show the uniqueness but I suppose you could just subscript them with numbers

Answer 15

ok.
so if we put the M first, as in
MWMWMWMWMWMW (1)
we have 6! 6! permutations
and we need to find that extra factor of 2 (or 4) or whatever it is to cover the other possibilities

Answer 16

it turns out that 6!6!*7 gives 3628800 , but I don;t see the logic of the 7 yet

Answer 17

the mm combination can show up in more than 2 places

Answer 18

Yeah^ I was writing them down I think you can find these possibilities:
\(W\color{red}{MM}WMWMWMWMW\)
\(WMW\color{red}{MM}WMWMWMW\)
\(WMWMW\color{red}{MM}WMWMW\)...

Answer 19

so 5 of those in addition to the original 2 sequences...

Answer 20

I think that will give 5
plus the 2 ways
with no mm combination

Answer 21

yah I think that's it :o

Answer 22

hmm, i was wondering about how to get those extras as well

Answer 23

that was indeed kind of tricky

Answer 24

if we label the combination WM =0 and the swapped pair MW=1
then we can start with
000000 (alternating W and M)
we are allowed one "swap" that gives us a MM adjacency
once we swap, we have to stay with the new order
thus,
000000
000001
000011
000111
001111
011111
111111

so 7 possibilities (times 6! 6!)

Answer 25

^Oh that's quite a nice way to see it

Answer 26

It's "after the fact" reasoning. And there must be a faster way to get to the solution.?!

Answer 27

My brain... It struggles to comprehend yet.

Answer 28

can you do part 1?

Answer 29

Not at all, lol.

Answer 30

Oh wait, I just got it. Alright. But Part 2 is the troublesome one.

Answer 31

If we treat each couple as one unit, then we can ask the question
how many ways to arrange 6 things:
6 in the first slot, 5 in the 2nd, etc
or 6*5*4*3*2*1 = 6!
if we swap the order of the first couple e.g. from MW to WM
we doubled the arrangements
notice we can swap the order of any of the 6 couples,
i.e. we can increase the 6! arrangements by 2*2*2*2*2*2 = 2^6
hence
6! 2^6

Answer 32

Part 2 is harder (for me,anyway)

we can break the problem into pieces.

the easier part:
assign 6 slots to M and 6 slots to W
now how many permutations can we have ?
the M can be assigned to their 6 slots 6! ways
the W can be assigned to their 6 slots 6! ways
multiply to get 6! 6! ways

*once we have decided which of the 12 slots is allowed to hold an M, and which a W*

Answer 33

second part:
how many ways can we assign 6 slots to be M (and of course the remaining 6 are for W) ?
with the requirement of *no adjacent W*

Answer 34

If this does not make sense, what part do you understand and what part is confusing ?

Answer 35

I assume you know this, but this is a useful summary (for easy problems!)
http://www.mathsisfun.com/combinatorics/combinations-permutations.html