Four female friends & four male friends will be pictured in a advertising photo. If the photographer wants to line them up in one row, with men & women alternating. How many possible arrangements may she chose?
A. 40320
B. 1680
C. 1152
D. 576
E. 70
OA is C
Could some expert pls elaborate on C, did some wrong calculations here.
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Four males can be arranged in 4! ways and four females can also be arranged in 4! waysRoland2rule wrote:Four female friends & four male friends will be pictured in a advertising photo. If the photographer wants to line them up in one row, with men & women alternating. How many possible arrangements may she chose?
A. 40320
B. 1680
C. 1152
D. 576
E. 70
OA is C
Could some expert pls elaborate on C, did some wrong calculations here.
With males on the left side of females the number of ways are 4! X 4! = 576.
With males on the right side of females the number of ways are 4! X 4! = 576.
=> Total number of ways = 576+576= 1152
Hence C is the correct option
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Hello Roland2rule.
Let's see your question.
We have to arrange 4 woman and 4 men.
We have 8 places: _ _ _ _ _ _ _ _. The possible arrangements are MWMWMWMW or WMWMWMWM.
The total possibles options are:
1) for MWMWMWMW = 4*4*3*3*2*2*1*1 = 4!*4! = 576.
2) for WMWMWMWM = 4*4*3*3*2*2*1*1 = 4!*4! = 576.
So, the answer total number is 576+576 = 1152.
This is why the correct answer is C.
I hope this explanation helps you.
I'm available if you'd like a follow-up.
Regards.
Let's see your question.
We have to arrange 4 woman and 4 men.
We have 8 places: _ _ _ _ _ _ _ _. The possible arrangements are MWMWMWMW or WMWMWMWM.
The total possibles options are:
1) for MWMWMWMW = 4*4*3*3*2*2*1*1 = 4!*4! = 576.
2) for WMWMWMWM = 4*4*3*3*2*2*1*1 = 4!*4! = 576.
So, the answer total number is 576+576 = 1152.
This is why the correct answer is C.
I hope this explanation helps you.
I'm available if you'd like a follow-up.
Regards.
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