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probability help

by mitaliisrani » Thu Apr 05, 2012 12:28 am
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 C
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by killer1387 » Thu Apr 05, 2012 12:51 am
mitaliisrani 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 C
there are 8 places.
Fix alternate 4 places You may want to start from first place by Female or first by male

hence 2 methods,
now you have two alternate sequence with 4 blanks and 4 female or male
hence its 4! *4!

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

hence C

Hope this helps..!!
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by Pharo » Thu Apr 05, 2012 12:55 am
The easiest way to solve this is by drawing:

Imagine the row starts from left with a male. You will have 4 people to choose from. Next a female. You have 4 people (4 females) to choose from. Next will be a male, 3 options (since you already chose 1). And female with 3 options etc.. So it will look like :

4-4-3-3-2-2-1-1

The number of possible arrangements from above is 4*4*3*3*2*2*1*1 = 576.

Now you are not done yet. What if the row starts with a female? It will be the same logic and outcome. So there are 576 additional possibilities.

The total answer is 576*2 = 1152 :)
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by ka_t_rin » Thu Apr 05, 2012 12:58 am
mitaliisrani 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 C
You can choose the 1st person in 8 ways, than you have a constraint - you should choose a person of another GMAT - 4 ways.
Now we have equal quantity of F&M = 3, so each two next people can be chosen in 3 ways. Then in two ways again two times, and finally we have 1 F and 1 M.

So the answer is 8x4x3x3x2x2x1x1=1152 - Choose C )))
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by shubham_k » Sun Apr 08, 2012 6:38 am
Lets say the photographer chooses male to stand first. So the first position can be chosen in 4 ways. Next position is for a female so again it acn be chosen in 4 ways. Next is a male position and can be chosen out of three remaining males so 3 ways for that. Similarly for next female position 3 ways.The remaining 2 position for males can be chosen in 2 ways. And similarly remaining two female positions can be chosen in 2 ways. So the total arrangement can be done in 4*3*2*4*3*2 = 4!*4!.

Now the photographer can alternatively choose female to stand first and following the above scenario arrangement can take place in 4!*4! ways. So total number of ways of arranging are 4!*4! + 4!*4!=2*4!*4!. Simple maths shows the answer is C :)


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by Anurag@Gurome » Sun Apr 08, 2012 5:45 pm
mitaliisrani 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 C
Men and women alternating can be done in the following 2 ways:

(1) M-W-M-W-M-W-M-W
(2) W-M-W-M-W-M-W-M

For each of the above cases, men can be arranged in 4! ways.
Similarly, women can be arranged in 4! ways.
Therefore, total number of arrangements = 2 * 4! * 4! = 2 * 4 * 3 * 2 * 4 * 3 * 2 = 1,152

The correct answer is C.
Anurag Mairal, Ph.D., MBA
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by factor26 » Tue Apr 10, 2012 2:25 pm
Was anyone else confused with the wording here? Initially I thought the ONLY way to organize was M W M W and my answer was 576. Anyone else see an issue with the wording here?
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