Last visit was: 28 Apr 2026, 20:49 It is currently 28 Apr 2026, 20:49
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Events & Promotions
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 28 Apr 2026
Posts: 109,950
Own Kudos:
811,823
 [7]
Given Kudos: 105,927
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,950
Kudos: 811,823
 [7]
Four representatives are to be randomly chosen from a committee of six 17 May 2021, 08:45
2
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

83% (02:00) correct 17% (02:12) wrong based on 182 sessions

History

Date
Time
Result
Not Attempted Yet
Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman?

A. 400
B. 415
C. 455
D. 480
E. 495
ShowHide Answer
Official Answer
D
User avatar
Niveditha28
User avatar
Joined: 08 Sep 2019
Last visit: 04 Sep 2022
Posts: 56
Own Kudos:
26
 [2]
Given Kudos: 101
GMAT 1: 600 Q43 V30
GMAT 1: 600 Q43 V30
Posts: 56
Kudos: 26
 [2]
Four representatives are to be randomly chosen from a committee of six 17 May 2021, 08:51
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Opt D - 12C4 -6C4 (total no.of 4 member team - no women team of 4)

480

Posted from my mobile device
avatar
RiyaJain69
avatar
Joined: 30 Dec 2020
Last visit: 30 Dec 2025
Posts: 116
Own Kudos:
69
Given Kudos: 112
Location: India
Schools: IIMC '25 (WA) IIMB '25 (D) IIMA PGPX'25 (D)
GMAT 1: 700 Q48 V39
Schools: IIMC '25 (WA) IIMB '25 (D) IIMA PGPX'25 (D)
GMAT 1: 700 Q48 V39
Posts: 116
Kudos: 69
Four representatives are to be randomly chosen from a committee of six 22 May 2021, 03:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I'm confused. Since one woman has to be in the chosen four, we have the following combinations for the remaining three positions:

1) 3W (5C3 = 10)
2) 2W + 1M (5C2 x 6C1 = 10 x 6 = 60)
3) 1W + 2M (5C1 x 6C2 = 5 x 15 = 75)
4) 3M (6C3 = 20)

The total is 165, so shouldn't this be the answer? Even if there is a shorter way to solve this, I need to understand why my approach is wrong.
User avatar
Nidzo
User avatar
Joined: 26 Nov 2019
Last visit: 02 Aug 2025
Posts: 958
Own Kudos:
1,480
 [1]
Given Kudos: 59
Location: South Africa
Posts: 958
Kudos: 1,480
 [1]
Four representatives are to be randomly chosen from a committee of six 22 May 2021, 03:29
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
RiyaJain69
I'm confused. Since one woman has to be in the chosen four, we have the following combinations for the remaining three positions:

1) 3W (5C3 = 10)
2) 2W + 1M (5C2 x 6C1 = 10 x 6 = 60)
3) 1W + 2M (5C1 x 6C2 = 5 x 15 = 75)
4) 3M (6C3 = 20)

The total is 165, so shouldn't this be the answer? Even if there is a shorter way to solve this, I need to understand why my approach is wrong.
Show more

Hey RiyaJain69, your numbers are wrong. Firstly, it is 4 representatives that need to be chosen while you were working out choosing 3. Secondly, there are 6 women and not 5.

In this case it is best to workout the possibilities of choosing no women, and subtracting it from the total possible ways in choosing 4 from the entire group of 12 (6 men and 6 women)
User avatar
chiplesschap
User avatar
Joined: 12 Jun 2023
Last visit: 25 Nov 2024
Posts: 47
Own Kudos:
21
Given Kudos: 43
Posts: 47
Kudos: 21
Four representatives are to be randomly chosen from a committee of six 31 Jan 2024, 11:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
While I get the 'total minus no woman' approach why am I not getting the same answer with the counting approach :

Case 1 : 4W+0M
-> (6 x 5 x 4 x 3)/(4!)

Case 2 : 3W+1M
-> (6 x 5 x 4 x 6)/4!

Case 3 : 2W+2M
-> (6 x 5 x 6 x 5)/4!

Case 4 : 1W+3M
-> (6 x 6 x 5 x 4)/4!

Where am I going wrong?

Thanks

Posted from my mobile device
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 28 Apr 2026
Posts: 109,950
Own Kudos:
811,823
 [1]
Given Kudos: 105,927
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,950
Kudos: 811,823
 [1]
Four representatives are to be randomly chosen from a committee of six 01 Feb 2024, 00:47
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chiplesschap
Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman?

A. 400
B. 415
C. 455
D. 480
E. 495

While I get the 'total minus no woman' approach why am I not getting the same answer with the counting approach :

Case 1 : 4W+0M
-> (6 x 5 x 4 x 3)/(4!)

Case 2 : 3W+1M
-> (6 x 5 x 4 x 6)/4!

Case 3 : 2W+2M
-> (6 x 5 x 6 x 5)/4!

Case 4 : 1W+3M
-> (6 x 6 x 5 x 4)/4!

Where am I going wrong?

Thanks

Posted from my mobile device
Show more

    4 women and 0 men = 6C4*1 = 6!/(4!2!) * 1 = 15

    3 women and 1 men = 6C3 * 6C1 = 6!/(3!3!) * 6 = 120

    2 women and 2 men = 6C2 * 6C2 = 6!/(2!2!) * 6!/(2!2!) = 225

    1 women and 3 men = 6C1 * 6C3 = 6 * 6!/(3!3!) = 120.

The sum of the above is 480.

Hope it helps.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,986
Own Kudos:
1,119
Posts: 38,986
Kudos: 1,119
Four representatives are to be randomly chosen from a committee of six 24 Mar 2025, 03:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109950 posts
Krunaal
Tuck School Moderator
852 posts