BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

probability help

Expert replies
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
Join the discussion
Source: — Problem Solving |

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..!!
Join the discussion

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 :)
Join the discussion

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 )))
Join the discussion

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 :)


If you feel this has been helpful to you please take a moment to click on thank.
Join the discussion

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
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

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?
Join the discussion