Permutation and combination question

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The discussion revolves around calculating the number of ways to position 5 men and 4 women for a photo shoot, with men arranged by height. The user initially calculated various groupings of women but arrived at a total of 2154, which differs from the expected answer of 3024. Key points include the need to accurately compute permutations and combinations for different group arrangements of women. Several participants pointed out errors in the user's calculations, particularly in the grouping cases. The conversation highlights the importance of careful consideration of all possible arrangements in combinatorial problems.
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Homework Statement


A photographer is positioning 5 men and 4 women for a photo shoot. The men are positioned in the order from shortest on the left to tallest on the right. Find the number of ways the photographer can position them in a row. (All men are of different heights and are not necessarily standing together in a group)

Homework Equations


The Attempt at a Solution


Since the position of men are fixed, i will just consider the possible permutation of women.

Four women can either stand together as one group, or split into two, three or four groups.

Below are combination forl 5 possible cases
Grouped together: 6 x 4! = 360
Form two groups of 2: 4C2/2 x 6P2 = 90
Form two groups of 1 and 3: 4 x 6P2 = 120
Form three groups of 1,1 and 2: 4 x 3 x 6P3 = 1440
All separated: 6P4 = 360

Then I added up all the possible combinations. The answer I obtained is 2154. However, the answer provided is 3024. Can anyone tell me which case I fail to consider? Or is there anything wrong with my working? Thanks
 
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gaobo9109 said:

Homework Statement


A photographer is positioning 5 men and 4 women for a photo shoot. The men are positioned in the order from shortest on the left to tallest on the right. Find the number of ways the photographer can position them in a row. (All men are of different heights and are not necessarily standing together in a group)


Homework Equations





The Attempt at a Solution


Since the position of men are fixed, i will just consider the possible permutation of women.

Four women can either stand together as one group, or split into two, three or four groups.

Below are combination forl 5 possible cases
Grouped together: 6 x 4! = 360
6x24=144, not 360.
Form two groups of 2: 4C2/2 x 6P2 = 90
I'm too tired to figure out what you're doing, but I got 360 for this case.
Form two groups of 1 and 3: 4 x 6P2 = 120
And 720 for this case.
Form three groups of 1,1 and 2: 4 x 3 x 6P3 = 1440
All separated: 6P4 = 360
We agree on these cases.
Then I added up all the possible combinations. The answer I obtained is 2154. However, the answer provided is 3024. Can anyone tell me which case I fail to consider? Or is there anything wrong with my working? Thanks
 

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