Last visit was: 10 Jul 2025, 10:58 It is currently 10 Jul 2025, 10:58
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,623
Own Kudos:
740,135
 [6]
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,623
Kudos: 740,135
 [6]
2
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
Chemerical71
Joined: 09 Jan 2016
Last visit: 01 Sep 2020
Posts: 77
Own Kudos:
447
 [2]
Given Kudos: 61
GPA: 3.4
WE:General Management (Human Resources)
Posts: 77
Kudos: 447
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
FB2017
Joined: 18 Jun 2017
Last visit: 14 Sep 2017
Posts: 50
Own Kudos:
Given Kudos: 165
Posts: 50
Kudos: 13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
Itisallinurhead
Joined: 15 Jan 2018
Last visit: 27 Dec 2018
Posts: 5
Own Kudos:
Given Kudos: 78
Posts: 5
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks in advance for any response!!!

why is the below method not correct?
A _ _ , A has to be one of the letters. For the remaining 2 position it will be --> 4c1*3c1=12.

Since, the correct answer is 6, the possible groups are:
ABC, ABD, ABE, ACD, ACE, ADE.
Each group can again be arranged in 3! ways. So, What would the approach/solution be if the question asked for number of arrangements?
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,819
Own Kudos:
5,867
 [2]
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,819
Kudos: 5,867
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Itisallinurhead
Thanks in advance for any response!!!

why is the below method not correct?
A _ _ , A has to be one of the letters. For the remaining 2 position it will be --> 4c1*3c1=12.

Since, the correct answer is 6, the possible groups are:
ABC, ABD, ABE, ACD, ACE, ADE.
Each group can again be arranged in 3! ways. So, What would the approach/solution be if the question asked for number of arrangements?

Hi Itisallinurhead

Here for the question being asked, in the 12 options possible when
you use 4c1*3c1 - AED and ADE are both possible(but since we are
asked to find the total possible groups) it would be wrong!

If the question read -
"total number of arrangements possible if A is one of the letters "
One of the letters will be A.
The other two letter can be chosen as 4c2 = 6
There are 3!(6) ways of arranging them.

Total arrangments possible are 1*6*6 = 36

Hope this helps you!
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 10 Jul 2025
Posts: 21,070
Own Kudos:
Given Kudos: 296
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,070
Kudos: 26,123
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
How many different groups of three distinct letters are possible if the three letters are chosen from the letters A, B, C, D, and E and A must be one of the letters selected?

A. 60
B. 10
C. 6
D. 4
E. 3

First, we know this will be a combination problem, since the order of selection doesn’t matter.

Since A must be selected, we are actually choosing only 2 letters from the 4 letters: B,C,D,E. Thus, the selection can be made in 4C2 = (4 x 3)/2 = 6 ways.

Answer: C
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,375
Own Kudos:
Posts: 37,375
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102623 posts
PS Forum Moderator
685 posts