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 →

How many combinations of three letters taken from letters

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri Sep 28, 2012 10:54 pm
gmatter2012 wrote:How many combinations of three letters taken from letters (a,a,b,b,c,c,d) are possible?

A. 12
B. 13
C. 35
D. 36
E. 56
Case 1: All 3 letters are different
Number of combinations of 3 that can be formed from 4 letters {a, b, c, d} = 4C3 = 4.

Case 2: 2 letters are the same
Number of options for the letter that appears twice = 3. (a, b, or c.)
Number of options for the third letter = 3. (Any of the other 3 letters.)
To combine these options, we multiply:
3*3 = 9.

Total options = 4+9 = 13.

The correct answer is B.

Some test-takers might find it easier simply to write out all of the possible combinations:

Case 1: All 3 letters are different
abc
abd
acd
bcd
4 options.

Case 2: 2 letters are the same
aab
aac
aad
bba
bbc
bbd
cca
ccb
ccd
9 options.

Total options = 4+9 = 13.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 138
Joined: Sat May 05, 2012 7:03 pm
Thanked: 1 times

by gmatter2012 » Sat Sep 29, 2012 12:31 am
GMATGuruNY wrote:
gmatter2012 wrote:How many combinations of three letters taken from letters (a,a,b,b,c,c,d) are possible?

A. 12
B. 13
C. 35
D. 36
E. 56
Case 1: All 3 letters are different
Number of combinations of 3 that can be formed from 4 letters {a, b, c, d} = 4C3 = 4.

Case 2: 2 letters are the same
Number of options for the letter that appears twice = 3. (a, b, or c.)
Number of options for the third letter = 3. (Any of the other 3 letters.)
To combine these options, we multiply:
3*3 = 9.

Total options = 4+9 = 13.

The correct answer is B.

Some test-takers might find it easier simply to write out all of the possible combinations:

Case 1: All 3 letters are different
abc
abd
acd
bcd
4 options.

Case 2: 2 letters are the same
aab
aac
aad
bba
bbc
bbd
cca
ccb
ccd
9 options.

Total options = 4+9 = 13.
Thank you Mitch

I was just confused as to the number of cases that need to be taken

case 1 : all are distinct 4C3= 4
case 2 : two of one kind and other different: 3C1*3C1= 9

where 3C1 choosing 1 which will form the pair from aa bb cc,
second 3C1 choosing 1 which will remain after we have chosen the pair .

lets say we have chosen aa for the pair then for the single letter we have bb, cc, and d so basically 3 distinct choices , for the single letter hence 3C1.

Thank you
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 47
Joined: Wed Sep 05, 2012 11:36 amFollowed by: 4 members
Thanked: 2 times

by eski » Sat Sep 29, 2012 2:45 am
Is this a right GMAT question and of what difficulty?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sat Sep 29, 2012 8:49 am
eski wrote:Is this a right GMAT question and of what difficulty?
I'm pretty sure that, in its current form, this would never be a true GMAT question. It's too ambiguous.
The main problem is the assumption that the word "combination" implies that the order of the letters does not matter. If this assumption is correct then Mitch's solution is perfect (as always).

Having said that, I don't think this is necessarily how people interpret the word "combination." For example, when you use a combination lock, the order of the numbers certainly matters.

As you can see in Mitch's solution, we have abc as one combination, but we're not including 5 other arrangements of the 3 letters (acb, bac, bca, cab, cba). Instead, we're just looking at the number of ways to select 3 letters, without considering their order.

To be a true GMAT question, any ambiguity must be removed.

One fix: In how many different ways can 3 letters be selected from {a,a,b,b,c,c,d} if the order in which the letters are selected does not matter. (I'm sure the test-maker could do better than this)

Alternatively, if the order does matter (i.e., abc, acb, bac, bca, cab and cba are considered different), then the wording could be: How many different 3-letter words can be constructed if the 3 letters are selected from {a,a,b,b,c,c,d}?

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1