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Smita04
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Four female friends & four male friends will be pictured in Updated on: 05 Apr 2012, 03:59
Originally posted by Smita04 on 05 Apr 2012, 01:30.
Last edited by Bunuel on 05 Apr 2012, 03:59, edited 1 time in total.
Edited the question and added the OA
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C
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45% (medium)

Question Stats:

63% (01:29) correct 37% (01:29) wrong based on 304 sessions

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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
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C
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Four female friends & four male friends will be pictured in 05 Apr 2012, 01:58
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Smita04
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
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okay you got 2 ways of assigning alternate places to either male or female

hence 2

now any of the alternate 4 places can be filled by 4 male or female in 4!
similarily,
other 4 alternate places can be filled in 4!

hence required probability= 2*4!*4!=1152

hence C

Hope this helps..!!
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Four female friends & four male friends will be pictured in 05 Apr 2012, 04:03
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Smita04
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
Show more

To meet the condition in the stem men and women should line up either: M-W-M-W-M-W-M-W or W-M-W-M-W-M-W-M. Now, for each of these two cases men can be arranged in 4! ways and women can be arranged also in 4! ways, hence total # of arrangements is 2*4!*4!=1,152.

Answer: C.

Hope it's clear.
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Four female friends & four male friends will be pictured in 31 Jan 2020, 07:48
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Smita04
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

To meet the condition in the stem men and women should line up either: M-W-M-W-M-W-M-W or W-M-W-M-W-M-W-M. Now, for each of these two cases men can be arranged in 4! ways and women can be arranged also in 4! ways, hence total # of arrangements is 2*4!*4!=1,152.

Answer: C.

Hope it's clear.
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Why can't it be 8!/4!*4! = 70?

Can you please explain?
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Bunuel
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Four female friends & four male friends will be pictured in 31 Jan 2020, 08:07
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Bunuel
Smita04
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

To meet the condition in the stem men and women should line up either: M-W-M-W-M-W-M-W or W-M-W-M-W-M-W-M. Now, for each of these two cases men can be arranged in 4! ways and women can be arranged also in 4! ways, hence total # of arrangements is 2*4!*4!=1,152.

Answer: C.

Hope it's clear.




Why can't it be 8!/4!*4! = 70?

Can you please explain?
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8!/(4!*4!) is the number ways to arrange 8 items out of which 4 are of type A and 4 are of type B. For example, the number of ways to arrange 33334444 is 8!/(4!4!).
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Four female friends & four male friends will be pictured in 27 Sep 2022, 16:27
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There is some problem with the answer i guess. Correct me if I am wrong.
SO we have 4 boys and 4 girls such that no two boys or no two girls sit togethe.
So we can do this question by gaping method: _B_B_B_B_
We have 5 places for four girls. So we can choose 4 places using 5C4.
Next we can arrange boys and girls in 4!*4! ways.

So the correct answer should be: 5C4*4!*4! = 2880.
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Four female friends & four male friends will be pictured in 18 Oct 2022, 13:55
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jim441
There is some problem with the answer i guess. Correct me if I am wrong.
SO we have 4 boys and 4 girls such that no two boys or no two girls sit togethe.
So we can do this question by gaping method: _B_B_B_B_
We have 5 places for four girls. So we can choose 4 places using 5C4.
Next we can arrange boys and girls in 4!*4! ways.

So the correct answer should be: 5C4*4!*4! = 2880.
Show more


5 places for 4 girls allows 1 place to be unfilled. That one place could be any of the spots between the boys, which would mean two boys sitting next to each other, violating the question stem.

Posted from my mobile device
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Four female friends & four male friends will be pictured in 18 Oct 2022, 14:56
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Regor60
jim441
There is some problem with the answer i guess. Correct me if I am wrong.
SO we have 4 boys and 4 girls such that no two boys or no two girls sit togethe.
So we can do this question by gaping method: _B_B_B_B_
We have 5 places for four girls. So we can choose 4 places using 5C4.
Next we can arrange boys and girls in 4!*4! ways.

So the correct answer should be: 5C4*4!*4! = 2880.


5 places for 4 girls allows 1 place to be unfilled. That one place could be any of the spots between the boys, which would mean two boys sitting next to each other, violating the question stem.

Posted from my mobile device
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Thanks a lot, this one my really messing up with me.
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Four female friends & four male friends will be pictured in 22 Dec 2025, 17:16
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Smita04
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
Show more

The word "arrangements" indicates permutations. There are a couple of ways to do this problem.

Method A: Permutations

There are 4 women and 4 men, and the pattern they're sitting in can be W M W ... or M W M.... So arrangement would be \(4!*4!*2=1152\).

Method B: Filling seats

You could also do this using counting principle, where you multiply together the options for each seat. This would result in:

\(4*4*3*3*2*2*1*1=576\)

But because there are two patterns (W M W or M W M), need to multiply by 2, which gives 1152.
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