# Permutation problem doesn't make sense

• Feb 21st 2010, 04:51 AM
iva
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 :

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

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!(Happy)
• Feb 21st 2010, 05:14 AM
Quote:

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 :

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

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!(Happy)

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.
• Feb 21st 2010, 05:19 AM
iva
Wow, put that way it makes sense now

THANK YOU !!!!
• Feb 21st 2010, 05:27 AM
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:)
• Feb 21st 2010, 06:04 AM
Quote:

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 \$\displaystyle (6!)^2=6!6!\$ arrangements.
• Feb 21st 2010, 09:24 AM
iva
Lovely, thank you so much!! I had just written out another whole problem and in doing so figured out where i went wrong!