A simple permutation question or not so simple

  • Thread starter johnnyICON
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How many ways can 2 men and 3 women be seated in a row such that no 2 men are sitting beside each other?

Now I have always had a problem with overthinking these kinds of questions. I'll usually write something down but then doubt myself.

What I did was simply did 2! * 3!.
3 * 2 * 2 * 1 * 1
W M W M W
That's by alternating women with men.

But here comes me overthinking the problem again.
I could altenate with men first, and then have 2 women sitting beside each other.
M W M W W right?
Or W W M W M....

Ahhh I feel like an idiot.
 

CRGreathouse

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mWWWm
mWWmW
mWmWW
WmWWm
WmWmW
WWmWm

In each of these arrangements, there are 2!3! ways to choose the men and women, so there are 72 ways to arrange them. (Unless, of course, I've made a mistake somewhere.)
 
79
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What if I were to up the ante of men and women to 4 and 5, respectively.
During an exam, I shouldn't be writing all the possible permutations for the problem.

I would get 4!5! for each arrangement, but how do I know how many arrangements there is going to be?
 
1,349
2
you must think of the 2 men as 1 with 2 permutation for MaMb and MbMa.
 
79
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I should of mentioed that each person is distinguishable from another. There are no two that are a like.
 
79
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What I got was,
mWmWmWmWW
WmWmWmWmW
WWmWmWmWm
mWWmWmWmW
WmWWmWmWm
mWmWWmWmW
WmWmWWmWm
mWmWmWWmW
WmWmWmWWm

There are 9 in total. Each arrangement is 5!4!.
Therefore 5!*4!*9 is the number of ways this can be done.

Is this coorect?
 
79
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here is a solution that I found, but uses a round table instead of a row.

How many ways can 5 man and 7 women be seated at a round table with
no 2 men next to each other?
Solution. First place the women in 6!. Now there are 7C5 ways to pick
5 spots for the men so that they are not adjancent. Finally, in each
of these 5 spots, the men can be placed in 5! ways. Hence, there are
5!6!7C5 = 1814400.

What I have is 5!4!5C4
 
509
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Do it this way:
First place the women in order. This is 5!. Then place the men in the spaces between the women. There are 6 spaces (including the ends) so you have C(6, 4) * 4! ways to place the men, and 5! * C(6, 4) * 4! altogether.

Now I'll check by listing:

110000 wwmwmwmwm
101000 wmwwmwmwm
100100 wmwmwwmwm
100010 wmwmwmwwm
100001 wmwmwmwmw
011000 mwwwmwmwm
010100 mwwmwwmwm
010010 mwwmwmwwm
010001 mwwmwmwmw
001100 mwmwwwmwm
001010 mwmwwmwwm
001001 mwmwwmwmw
000110 mwmwmwwwm
000101 mwmwmwwmw
000011 mwmwmwmww

So there are indeed C(6, 2) = 15 ways to arrange the men and women before permuting.
 
79
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BicycleTree said:
Do it this way:
First place the women in order. This is 5!. Then place the men in the spaces between the women. There are 6 spaces (including the ends) so you have C(6, 4) * 4! ways to place the men, and 5! * C(6, 4) * 4! altogether.

Now I'll check by listing:

110000 wwmwmwmwm
101000 wmwwmwmwm
100100 wmwmwwmwm
100010 wmwmwmwwm
100001 wmwmwmwmw
011000 mwwwmwmwm
010100 mwwmwwmwm
010010 mwwmwmwwm
010001 mwwmwmwmw
001100 mwmwwwmwm
001010 mwmwwmwwm
001001 mwmwwmwmw
000110 mwmwmwwwm
000101 mwmwmwwmw
000011 mwmwmwmww

So there are indeed C(6, 2) = 15 ways to arrange the men and women before permuting.

awesome man, this is exactly how my prof wanted us to do it. thanks
 
230
0
johnnyICON said:
How many ways can 2 men and 3 women be seated in a row such that no 2 men are sitting beside each other?
Don't forget you can always have two of the women sit on the men's lap. :!!)
 
Last edited:

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