# Thread: Permutation problem doesn't make sense

1. ## Permutation problem doesn't make sense

Hi there, I have this problem in my discreet maths text book. The answer is given but my answer was half the value and i don't understand why my answer needs to be double. Any enlightment would be most appreciated :

A group contains 6 men and 6 women. How many ways are there to
arrange these people in a row if the men and women alternate.

My solution :

I see 2 ways that this works, either start with men:
m w m w m w m w m w m w

or start with women:

w m w m w m w m w m w m

there are 6! permutations for the first group and also 6! for the 2nd group . So i get the overall total number of ways to arrange men and women alternating is 6!6! = 518400 but the answer is actually 6!6!2 ,why double the answer??

Many thanks!

2. Originally Posted by iva
Hi there, I have this problem in my discreet maths text book. The answer is given but my answer was half the value and i don't understand why my answer needs to be double. Any enlightment would be most appreciated :

A group contains 6 men and 6 women. How many ways are there to
arrange these people in a row if the men and women alternate.

My solution :

I see 2 ways that this works, either start with men:
m w m w m w m w m w m w

or start with women:

w m w m w m w m w m w m

there are 6! permutations for the first group and also 6! for the 2nd group . So i get the overall total number of ways to arrange men and women alternating is 6!6! = 518400 but the answer is actually 6!6!2 ,why double the answer??

Many thanks!
Hi iva,

Women first... WM.. WM . WM.. WM.. WM.. WM
.................... 6(6). 5(5). 4(4). 3(3). 2(2). 1(1) = 6!6!

men first...same calculation, giving (2)6!6!

Any of the 6 women can be first, followed by any of the 6 men.
Any of the remaining 5 women can be next, followed by any of the remaining 5 men, and so on.

3. Wow, put that way it makes sense now

THANK YOU !!!!

4. Correct me if i'm wrong, but in this next case isn't the answer 6!6!:

if all the woman are arranged in a row IN FRONT of all the men. women ALWAYS in front . then the combinations of these would surely be 6! for the men row and 6! for women, and overall 6!6! arrangements?

Thanks again

5. Originally Posted by iva
Correct me if i'm wrong, but in this next case isn't the answer 6!6!:

if all the woman are arranged in a row IN FRONT of all the men. women ALWAYS in front . then the combinations of these would surely be 6! for the men row and 6! for women, and overall 6!6! arrangements?

Thanks again
Hi again iva,

yes, you are right.

For any single row of women, the men can be arranged in 6! ways.

As there are 6! rows of women, then there are $(6!)^2=6!6!$ arrangements.

6. Lovely, thank you so much!! I had just written out another whole problem and in doing so figured out where i went wrong!