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
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Letter combination

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2789
Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members
Location: Chennai, India
Thanked: 206 times
GMAT Score:640
by GmatKiss » Mon Aug 15, 2011 12:25 am
How many words can be formed with the letters of the word 'OMEGA' when,

(i) 'O' and 'A' occupying end places.

(ii) 'E' being always in the middle

(iii) Vowels occupying odd-places

(iv) Vowels being never together.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members
Location: London
Thanked: 70 times

by kmittal82 » Mon Aug 15, 2011 12:42 am
(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

(2)
[1] [2] E [3] [4]

[1] can be filled in 4 ways, [2] in 3 ways, [3] in 2 ways , total ways = 4x3x2 = 24

(3)
[V1] [C1] [V2] [C2] [V3] where [V*] is Vowel, [C*] is consonant

V1 can be filled in 3 ways
C1 can be filled in 2 ways
V2 can be filled in 2 ways
C2 can be filled in 1 way
V3 can be filled in 1 way

total ways = 3 x 2 x 2 x 1 x 1 = 12


(4)
Vowels must occupy odd places for this, so same as (3) i.e. 12

OAs please?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Sun Jul 03, 2011 7:13 am
Thanked: 2 times

by Options » Mon Aug 15, 2011 5:36 am
kmittal82 wrote:(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

I think you should times another 2! to 6 since O and A can swap places. The qn said at the ends but did not specify which letter at which end.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members
Location: London
Thanked: 70 times

by kmittal82 » Mon Aug 15, 2011 5:40 am
Options wrote:
kmittal82 wrote:(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

I think you should times another 2! to 6 since O and A can swap places. The qn said at the ends but did not specify which letter at which end.
ah, thanks for correcting that, agreed, it ought to be 12
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Tue Aug 30, 2011 5:04 am
Thanked: 1 times

by viv09 » Tue Aug 30, 2011 7:07 pm
Hi. can someone please explain part 4 of the question

I believe for this. we can do the below.
Arrange such that vowels are always together

no of ways the letters can be arranges = 5!

no of ways vowels always together = 2x1x3! = 12
So vowels not together = 120 - 12 = 108..

please correct me if i am wrong

Sorry for the typo earlier.. had to edit
Last edited by viv09 on Tue Aug 30, 2011 7:32 pm, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Wed May 25, 2011 5:02 pm

by eduardo_cruza » Tue Aug 30, 2011 7:27 pm
i) O and A occupying end places = 12 ways
ii) E being always at the middle = 24 ways
iii) vowels occupying odd places = 12 ways
iv) vowels being never together = 5! - 3!x3 = 102 ways
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 608
Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members
Thanked: 37 times

by saketk » Wed Aug 31, 2011 11:09 am
viv09 wrote:Hi. can someone please explain part 4 of the question

I believe for this. we can do the below.
Arrange such that vowels are always together

no of ways the letters can be arranges = 5!

no of ways vowels always together = 2x1x3! = 12
So vowels not together = 120 - 12 = 108..

please correct me if i am wrong

Sorry for the typo earlier.. had to edit
Hi-- Your approach is correct but why did you multiply 2 in the part I marked RED?

The answer to 4th part: number of alphabets = 5
so they can be arranged in 5! ways. Now we have a constraint that vowels cannot be together.
So let's find out the cases when vowels are always together-- total vowels 3 (O,E,A) - treat them as 1 letter. so now we have 3 letter M,G & (OEA) -- they can be arranged in 3! ways. Also the 3 vowels can be arranged among themselves in 3! ways.. So total number of ways = 3!*3! = 36

So, our answer will be 5! - 3!*3! = 120- 36 = 84 ways.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 608
Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members
Thanked: 37 times

by saketk » Wed Aug 31, 2011 11:12 am
kmittal82 wrote:(1)
Fixing O and A, we have this O [1] [2] [3] A where [1] [2] and [3] represent blank spaces

[1] can be filled in 3 ways, [2] in 2 ways, so total ways = 6

(2)
[1] [2] E [3] [4]

[1] can be filled in 4 ways, [2] in 3 ways, [3] in 2 ways , total ways = 4x3x2 = 24

(3)
[V1] [C1] [V2] [C2] [V3] where [V*] is Vowel, [C*] is consonant

V1 can be filled in 3 ways
C1 can be filled in 2 ways
V2 can be filled in 2 ways
C2 can be filled in 1 way
V3 can be filled in 1 way

total ways = 3 x 2 x 2 x 1 x 1 = 12


(4)
Vowels must occupy odd places for this, so same as (3) i.e. 12

OAs please?
Hi -- your answer to 4th part is incorrect.. It should be 5!-3!*3! = 84 ways.
why do you think that vowels must occupy odd places?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Tue Aug 30, 2011 2:40 am

by Krk » Wed Aug 31, 2011 11:52 am
Hi -- your answer to 4th part is incorrect.. It should be 5!-3!*3! = 84 ways.
why do you think that vowels must occupy odd places?
Since this is a five letter word 1 2 3 4 5.
If vowels occupy even places, then there are only 2 places where vowels can be used.
Since there are 3 vowels, vowels need to take odd places only.
Hence, answer for 3rd and 4th question has to be same.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1