Last visit was: 22 May 2025, 12:14 It is currently 22 May 2025, 12:14
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
gmatt1476
Joined: 04 Sep 2017
Last visit: 27 Mar 2025
Posts: 334
Own Kudos:
Given Kudos: 62
Posts: 334
Kudos: 23,837
 [121]
7
Kudos
Add Kudos
113
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 22 May 2025
Posts: 8,213
Own Kudos:
4,715
 [21]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy)
Products:
GMAT Focus 1: 545 Q79 V79 DI73
Posts: 8,213
Kudos: 4,715
 [21]
17
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
UkrHurricane
Joined: 04 Dec 2017
Last visit: 28 Feb 2025
Posts: 50
Own Kudos:
56
 [21]
Given Kudos: 429
Location: Ukraine
Concentration: Finance, Human Resources
Schools: Simon '27
GMAT 1: 580 Q38 V32
GMAT 2: 590 Q49 V23
GMAT 3: 580 Q43 V27
GMAT 4: 630 Q45 V31 (Online)
GMAT 5: 670 Q47 V35 (Online)
GPA: 3.4
Schools: Simon '27
GMAT 5: 670 Q47 V35 (Online)
Posts: 50
Kudos: 56
 [21]
17
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
General Discussion
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,828
Own Kudos:
5,795
 [9]
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,828
Kudos: 5,795
 [9]
5
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Let the remaining contestants be A,B,C,D, and E.

They are listed as follows
ABCD | ABCE | BCDE | ACDE | ABDE

Therefore, there are \(C_4^{5} = 5\)(Option A) possible combinations for the semifinalists.
avatar
icanhazmba
Joined: 01 Jun 2019
Last visit: 24 Sep 2022
Posts: 76
Own Kudos:
Given Kudos: 40
Location: United States
Concentration: Strategy
GMAT 1: 740 Q49 V41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel, could you please explain? Thanks.
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 22 May 2025
Posts: 5,597
Own Kudos:
5,025
 [3]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,597
Kudos: 5,025
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gmatt1476
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

A. 5
B. 6
C. 7
D. 14
E. 21


PS07602.01

Given:Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected.

Asked: Of the different possible selections, how many contain neither Ben nor Ann?

There are 5 other participants out of which 4 are to be selected

Number of ways = 5C4 = 5

IMO A

Posted from my mobile device
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,791
Own Kudos:
12,379
 [10]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,791
Kudos: 12,379
 [10]
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Hi All,

We're told that Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. We're asked of the different possible selections, how many contain NEITHER Ben nor Ann. Since we're selecting 'groups' of people, this is ultimately a Combination Formula question.

IF we had NO restrictions, then with 7 contestants, the number of 'groups of four' would be 7c4 = 7!/4!3! = (7)(6)(5)/(3)(2)(1) = 35 possible groups of four

However, we're interested in groups that contain NEITHER Ben nor Ann, which means that we have just 5 people to choose from. 5c4 = 5!/4!/1! = (5)/(1) = 5 groups of four.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,761
Own Kudos:
33,641
 [10]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,761
Kudos: 33,641
 [10]
5
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
gmatt1476
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

A. 5
B. 6
C. 7
D. 14
E. 21


PS07602.01

Let Ann, Bob, C, D, E, F, G be the 7 contestants.

To ensure that neither Ben nor Ann are among the four semifinalists, let's remove them from the list of contestants.

So, we can select the four semifinalists from {C, D, E, F, G}

Since the order in which we select the 4 semifinalists does not matter, we can use COMBINATIONS.
We can select 4 semifinalists from 5 contestants in 5C4 ways.
5C4 = 5

Answer: A

Cheers,
Brent
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 22 May 2025
Posts: 20,812
Own Kudos:
25,879
 [1]
Given Kudos: 292
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: 20,812
Kudos: 25,879
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gmatt1476
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

A. 5
B. 6
C. 7
D. 14
E. 21


PS07602.01

If neither Ben nor Ann is among the semifinalists, then there are only 5 contestants vying for the four slots. Thus, the number of ways to select the semifinalists when neither Ben nor Ann is selected is 5C4 = 5.

Answer: A
avatar
Anki3188
Joined: 07 Sep 2021
Last visit: 15 May 2022
Posts: 1
Given Kudos: 7
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
It might be very silly but I have done it like 7C4- 2C2*5C2.
Total- ways of choosing Ben and Ann and * ways of choosing the remaining. What is the conceptual mistake I am making?
Bunuel chetan2u
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,791
Own Kudos:
12,379
 [1]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,791
Kudos: 12,379
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Anki3188,

Since the question is asking for the number of groups that do NOT include Ben AND do NOT include Ann, then your calculation would need to remove any group that includes Ben OR Ann OR both of them. Your calculation appears to only remove the groups that include both Ben and Ann.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]
avatar
Argp
Joined: 10 Aug 2019
Last visit: 30 May 2022
Posts: 44
Own Kudos:
Given Kudos: 151
Location: India
Posts: 44
Kudos: 10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u Bunuel I know I am coming up with the wrong answer, but can you help me understand what is wrong with the below approach -
[Total number of ways to select 4 candidates out of 7] - [number of ways Ben and Ann can be selected in the semifinalists]

Total number of ways to select 4 candidates out of 7 = 7C4 = 35

number of ways Ben is selected in the semifinalists (except Ann) = B_ = 1x5 ways = 5
number of ways Ann is selected in the semifinalists (except Ben) = A_ = 1x 5 ways = 5
ben and ann together as semifinalists = 1 way
number of ways Ben and Ann can be selected in the semifinalists => 5+ 5 + 1 = 11 ways

=> 35 - 11 = 24 ways
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 19 May 2025
Posts: 11,307
Own Kudos:
40,718
 [1]
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,307
Kudos: 40,718
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Argp
chetan2u Bunuel I know I am coming up with the wrong answer, but can you help me understand what is wrong with the below approach -
[Total number of ways to select 4 candidates out of 7] - [number of ways Ben and Ann can be selected in the semifinalists]

Total number of ways to select 4 candidates out of 7 = 7C4 = 35

number of ways Ben is selected in the semifinalists (except Ann) = B_ = 1x5 ways = 5
number of ways Ann is selected in the semifinalists (except Ben) = A_ = 1x 5 ways = 5
ben and ann together as semifinalists = 1 way
number of ways Ben and Ann can be selected in the semifinalists => 5+ 5 + 1 = 11 ways

=> 35 - 11 = 24 ways

You have gone wrong in your calculations

number of ways Ben is selected in the semifinalists (except Ann) = B_ = 1x5C3 ways = 10....You have to select 3 from remaining 5.
number of ways Ann is selected in the semifinalists (except Ben) = A_ = 1x 5 ways = 5...10 ways as above
ben and ann together as semifinalists = 1 way .......2C2*5C2=10
number of ways Ben and Ann can be selected in the semifinalists => 5+ 5 + 1 = 11 ways.......10+10+10=30 ways

Total = 35-30=5

Anki3188 , you have gone wrong in not taking cases where only one of them is selected. The answer would be correct if the question were to find ways in which both are not there.
User avatar
TalonShade
Joined: 21 Apr 2021
Last visit: 29 Dec 2022
Posts: 31
Own Kudos:
Given Kudos: 26
Location: India
GRE 1: Q165 V163
GPA: 2.24
GRE 1: Q165 V163
Posts: 31
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I am trying to do this using the counting method and would appreciate some help where I'm going wrong. I got the correct answer using formulas but cant seem to with the counting method.

So basically

Total ways of selecting 4 of 7 will give us all the ways - 7x6x5x4

And subtracting the following should give us the desired answer

1. Where Ben is a semifinalist but Ann is not - 1 x 5 x 4 x 3 ----- (a)
2. Where Ann is a semifinalist but Ben is not - 1 x 5 x 4 x 3 ----- (b)

T -(a+b) = 840 -(60+60) = 740 --- ???
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 19 May 2025
Posts: 11,307
Own Kudos:
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,307
Kudos: 40,718
Kudos
Add Kudos
Bookmarks
Bookmark this Post
TalonShade
I am trying to do this using the counting method and would appreciate some help where I'm going wrong. I got the correct answer using formulas but cant seem to with the counting method.

So basically

Total ways of selecting 4 of 7 will give us all the ways - 7x6x5x4

And subtracting the following should give us the desired answer

1. Where Ben is a semifinalist but Ann is not - 1 x 5 x 4 x 3 ----- (a)
2. Where Ann is a semifinalist but Ben is not - 1 x 5 x 4 x 3 ----- (b)

T -(a+b) = 840 -(60+60) = 740 --- ???


It is selection and you are taking arrangements.

So T = 7C4 = 35

a) B is one, so choose remaining 3 from available 5( less A and B) = 5C3 = 10
b) A and n(B): Similarly 10 ways for A being a part and B not being a part.
c) both (A and B): Also add ways where both A and B are part, so choose 2 out of remaining 5 = 5C2 = 10
Our answer = 35-10-10-10 = 5
A

There are various combinations in which above 3 could be calculated.
It could be (T-(A as one)-(B there but not A))=35-6C3-5C3=35-20-10=5
OR T-(A)-(B)+both(A and B)=35-20-20+10=5
User avatar
Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 443
Own Kudos:
Given Kudos: 5,901
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
GMAT 1: 450 Q42 V34
Posts: 443
Kudos: 41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
gmatt1476
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

A. 5
B. 6
C. 7
D. 14
E. 21


PS07602.01

Let Ann, Bob, C, D, E, F, G be the 7 contestants.

To ensure that neither Ben nor Ann are among the four semifinalists, let's remove them from the list of contestants.

So, we can select the four semifinalists from {C, D, E, F, G}

Since the order in which we select the 4 semifinalists does not matter, we can use COMBINATIONS.
We can select 4 semifinalists from 5 contestants in 5C4 ways.
5C4 = 5

Answer: A

Cheers,
Brent

Hi BrentGMATPrepNow, not sure why if I calculate all contestants first then minus Ben and Ann is not correct?

7C4 = 35
5C4 = 5
35-5 = 30
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,761
Own Kudos:
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,761
Kudos: 33,641
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kimberly77

Hi BrentGMATPrepNow, not sure why if I calculate all contestants first then minus Ben and Ann is not correct?

7C4 = 35
5C4 = 5
35-5 = 30

If you want to go that route you need to follow the property that says:
Number of ways to follow the restriction = (number of ways to ignore the restriction) - (number of ways to break the restriction)

The restriction here is that neither Ben nor Ann can be in the group of 4 semi finalists.

So we get:
Number of outcomes in which neither Ben nor Ann are in the group of 4 semi finalists = (number of 4-person groups chosen from from all seven semi-finalists) - (number of ways to break the restriction)

Here, your calculation of 5C4 = 5 does not give us the total number of ways to BREAK the restriction.
User avatar
Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 443
Own Kudos:
Given Kudos: 5,901
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
GMAT 1: 450 Q42 V34
Posts: 443
Kudos: 41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
Kimberly77

Hi BrentGMATPrepNow, not sure why if I calculate all contestants first then minus Ben and Ann is not correct?

7C4 = 35
5C4 = 5
35-5 = 30

If you want to go that route you need to follow the property that says:
Number of ways to follow the restriction = (number of ways to ignore the restriction) - (number of ways to break the restriction)

The restriction here is that neither Ben nor Ann can be in the group of 4 semi finalists.

So we get:
Number of outcomes in which neither Ben nor Ann are in the group of 4 semi finalists = (number of 4-person groups chosen from from all seven semi-finalists) - (number of ways to break the restriction)

Here, your calculation of 5C4 = 5 does not give us the total number of ways to BREAK the restriction.

Thanks BrentGMATPrepNow, in 5C4 = 5 which I have already removed Ben and Ann from 7 therefore 5 contestants left, then select 4 out of 5 semi-finalists. So not sure why this still doesn't include the total number of ways to BREAK the restriction? Could we not remove both Ben and Ann at the same time since question asked neither Ben nor Ann are to be in semi-finalists?
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,761
Own Kudos:
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,761
Kudos: 33,641
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kimberly77
Thanks BrentGMATPrepNow, in 5C4 = 5 which I have already removed Ben and Ann from 7 therefore 5 contestants left, then select 4 out of 5 semi-finalists. So not sure why this still doesn't include the total number of ways to BREAK the restriction? Could we not remove both Ben and Ann at the same time since question asked neither Ben nor Ann are to be in semi-finalists?


The restriction is that the 4-person selection CANNOT contain Ben or Ann.
Since 5C4 is the total number of 4-person selections that do NOT contain Ben or Ann, then you aren't breaking the restriction; you're obeying the restriction.
User avatar
Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 443
Own Kudos:
Given Kudos: 5,901
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
GMAT 1: 450 Q42 V34
Posts: 443
Kudos: 41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
Kimberly77
Thanks BrentGMATPrepNow, in 5C4 = 5 which I have already removed Ben and Ann from 7 therefore 5 contestants left, then select 4 out of 5 semi-finalists. So not sure why this still doesn't include the total number of ways to BREAK the restriction? Could we not remove both Ben and Ann at the same time since question asked neither Ben nor Ann are to be in semi-finalists?


The restriction is that the 4-person selection CANNOT contain Ben or Ann.
Since 5C4 is the total number of 4-person selections that do NOT contain Ben or Ann, then you aren't breaking the restriction; you're obeying the restriction.

Thanks BrentGMATPrepNow, think I get it now.
So we need to calculate 3 cases of no Ben 5C3, no Ann 5C3, no Ben & Ann 2C2 * 5C2. Total 35-30 = 5

Mistake I made here was I interpret neither Ben nor Ann as both not on there. Therefore only calculate it once. How will I know to interpret neither Ben nor Ann as either or? Thanks Brent
 1   2   
Moderators:
Math Expert
101653 posts
PS Forum Moderator
585 posts