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by BTGmoderatorRO » Wed Dec 13, 2017 4:35 pm
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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by GMATWisdom » Thu Dec 14, 2017 5:21 am
Roland2rule 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.
Four males can be arranged in 4! ways and four females can also be arranged in 4! ways
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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by EconomistGMATTutor » Fri Dec 15, 2017 8:34 am
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.
GMAT Prep From The Economist
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