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.

Combination

Expert replies
by sam2304 » Fri Jun 01, 2012 6:49 pm
A Â…firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
https://gmatandbeyond.blogspot.in/
Join the discussion
Source: — Problem Solving |

by eagleeye » Fri Jun 01, 2012 7:04 pm
Hi sam2304:

We have 4 departments with 4 people in each. To select 3 members, each one from a distinct team, we need to select 1 person each from 3 of the teams.

Step 1: Choose 3 teams out of 4. No of ways = 4C3 = 4
Step 2: Choose 1 person each from each of the selected teams. No of ways = 4C1*4C1*4C1 = 4*4*4
Total ways = (Choose 3 teams out of 4)*(select one person each of the 3 teams) = [spoiler]4*4*4*4 = 4^4. Hence B[/spoiler].

Another way: (By permutation and then combination)

There are total 16 people. We can select the first person from 16, then the second from 12, and the third from remaining 8. (Since we can only select 1 from each department).
So, no of ways = 16*12*8. Now the order of selection of those people is not important. Hence, we divide by 3!

Total no. of ways = (16*12*8)/3!= [spoiler]4^4. Hence still B[/spoiler]

Let me know if this helps :)
Join the discussion

by sam2304 » Fri Jun 01, 2012 11:42 pm
OA : B I solved using the first method posted by you, but wasn't sure whether it was the right approach. The doubt is cleared now :)
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
https://gmatandbeyond.blogspot.in/
Join the discussion