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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

GMATPrep Problem .. assign identification

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members
Thanked: 146 times

Re: GMATPrep Problem .. assign identification

by parallel_chase » Thu Sep 11, 2008 8:59 pm
shahab03 wrote:first of all is this permutation or combination problem. also if you can help me understand. thanks



Image

I guess the answer should be 4536.

This problem is of permutations.

We have total 10 letters [0,1,2,3,4,5,6,7,8,9]

first place can be filled in 9 ways, since 0 is not to be included = 9 ways.
Second place can also be filled in 9 ways, since 0 can be included = 9 ways.
Third place can be filled in 8 ways
Fourth place can be filled in 7 ways.

9*9*8*7 = 4536

Whats the OA?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 66
Joined: Sun Jun 29, 2008 8:37 pm
Location: India
Thanked: 1 times

by msd_2008 » Thu Sep 11, 2008 9:18 pm
How can the answer be 4536?
I would go about this problem in the following way:

1st digit can be arranged in 9 ways (0 not included)
2nd digit can be arranged in 10 ways (0 included)
3rd digit can be arranged in 10 ways (0 included)
4th digit can be arranged in 10 ways (0 included)

Hence total ways = 9*10*10*10 = 9000 ways

We cannot assume that same numbers should not be repeated in different digit places because thats nowhere mentioned in the problem.
If we assume this, then 4536 is the right answer.
Can you please let us know the OA?

Regards
MSD
When the going gets tough, the tough gets going.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members
Thanked: 146 times

by parallel_chase » Thu Sep 11, 2008 9:22 pm
msd_2008 wrote:How can the answer be 4536?
I would go about this problem in the following way:

1st digit can be arranged in 9 ways (0 not included)
2nd digit can be arranged in 10 ways (0 included)
3rd digit can be arranged in 10 ways (0 included)
4th digit can be arranged in 10 ways (0 included)

Hence total ways = 9*10*10*10 = 9000 ways

We cannot assume that same numbers should not be repeated in different digit places because thats nowhere mentioned in the problem.
If we assume this, then 4536 is the right answer.
Can you please let us know the OA?

Regards
MSD
The question explicitly states that "Each number consists of four different digits"

No Assumptions!!

Hope this helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Fri Sep 19, 2008 10:02 pm
Shahab,
This is a permutaion since the order does matter.

Eg: 7586 id is different from 7856 id (even though they have the same digits but the position of the digits make the difference)

Make 4 slots

1st slot can be filled by 9 numbers(since 0 cannot be included)
2nd slot by 9(the first number u have choosen cannot be included but 0 can be the 2nd digit)
3rd slot can be filled by 8 since we cant use the first 2 numbers in the first 2 slots
4th slot by 7 (same reason as above)

So 9*9*8*7 would be 4536
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Fri Sep 19, 2008 10:10 pm
Easier way to understand this would be

Take 0,1,2 How many different 3 digit identifcation numbers can be formed where 0 cannot be the first digit(since its 012 is not valid)

0,1,2

Brute force listing of possibilities:
102
120
201
210

4 numbers


Permutation/counting principle method:

1st slot: 2
2nd slot: 2
3rd slot:1

2*2*1 = 4

Hope this helps!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1