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
GMATBootcamp Starts Sep 21
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE BOOTCAMP

Live Online Bootcamp Class with Top GMAT Expert Chris Peckover

Sep 21 to Oct 9, 2026

Schedule
Mon to Fri · 7:00 to 10:00 PM ET
Included
Live classes + 6 months of TTP OnDemand
  • Boost your GMAT score in less than one month in a live online class
  • 6 months access to TTP OnDemand video courses included
View bootcamp & 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.

Need approach for this problem

Expert replies
by Tryingmybest » Tue Mar 24, 2009 11:05 am
I came across this question in Princeton . Unfortunately I reset the test. Please help me with the approach to this problem.

What is the probability of choosing four numbers 1, 2, 3, and 4 by four different persons A,B , C and D such that every Person chooses a distinct number?

I dont have a OA for this problem. But would like to know the approach.
Thanks!
Join the discussion
Source: — Problem Solving |

by devesh99 » Tue Mar 24, 2009 12:56 pm
Just trying

For selection of distinct numbers:
A can choose any of the 4 numbers (1,2,3,4) = 4
B can choose from any of the 3 remaining numbers = 3
C can choose from any of the 2 remaining numbers = 2
D can choose the last remaining number = 1
Therefore, 4*3*2*1 = 24

Now, for selection of numbers
A can select any of the 4 numbers = 4
B can select any of the 4 numbers = 4
C can select any of the 4 numbers = 4
D can select any of the 4 numbers = 4
Total chances = 4*4*4*4

Therefore Prability = 24/26 = 3/32
Join the discussion

by lilu » Tue Mar 24, 2009 3:13 pm
devesh99 wrote:Just trying



Therefore Prability = 24/26 = 3/32
I am sure you meant 24/246 :)
Join the discussion

by Tryingmybest » Tue Mar 24, 2009 4:11 pm
Thanks, I guess you are right. I remember an answer choice 10%

Appreciate your response.
Join the discussion