Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart 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

An interesting probability problem

Expert replies
by billzhao » Thu Feb 05, 2009 11:57 pm
Find the number of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.

My thinking is: I put John and Susan together as (JS) or (SJ)
there are eight categories of combinations as below:

(JS)BGBGBG (SJ)GBGBGB
BG(JS)BGBG GB(SJ)GBGB
BGBG(JS)BG GBGB(SJ)GB
BGBGBG(JS) GBGBGB(SJ)

So the number of combination is: 2*4!*4! - 8*3!*3! Is it correct?

The answer is 2*4!*4! - 14*3!*3!

Thanks.
Yiliang
Join the discussion
Source: — Problem Solving |

by fleurdelisse » Fri Feb 06, 2009 1:44 am
You under-counted the number of times john and susan can be next to each other, it is 14. 7 for when you start the seating with a boy and 7 for when you start the seating with a girl.

you counted in pairs of (1,2), (3,4), etc. and did not account for susan and john to be in seats (2,3) for example. Do the counting again, and you'll see what I mean.

Reasoning for the whole answer, for the other readers:

Total number of ways that 4 boys an 4 girls can be seated alternatively is:
2*4!*4!, since you have 4! for seating girls and 4! for seating boys, and you multiply by 2 to account for the fact that the first seat can be either a boy or a girl

and then you remove the possibilities of having John and Susan together, which is in 14 different ways (count only for when a boy is in the first seat for example, you'll get 7 and then multiply by 2). And for each of the 14 different ways, you can seat the remaining girls and boys in 3!*3! ways

So end result: 2*4!*4! - 14*3!*3!
Join the discussion

by fleurdelisse » Fri Feb 06, 2009 1:47 am
so, following your method of counting, it would be like this (in bold are those you did not account for):

(JS)BGBGBG (SJ)GBGBGB
B(SJ)GBGBG G(JS)BGBGB
BG(JS)BGBG GB(SJ)GBGB
BGB(SJ)GBG GBG(JS)BGB
etc.
Join the discussion

by billzhao » Fri Feb 06, 2009 3:20 am
i got it. thanks a lot!
Yiliang
Join the discussion

by Uri » Fri Feb 06, 2009 3:23 pm
Is there any other way to find out the solution, without writing down the possible arrangements?
Join the discussion