NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 01:13 It is currently 26 Apr 2024, 01:13
Decision Tracker
My Rewards
New posts
New comers' posts

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

A certain company consists of 3 managers and 8 non-managers.

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
B
galiya
Manager
Manager
Joined: 16 Jan 2011
Posts: 72
Own Kudos [?]: 715 [11]
Given Kudos: 15
Send PM
A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   26 Apr 2012, 12:24
1
Kudos
10
Bookmarks
00:00
A
B
C
D
E

Difficulty:

35% (medium)

Question Stats:

74% (02:09) correct 26% (02:33) wrong based on 288 sessions
Hide Show timer Statistics
A certain company consists of 3 managers and 8 non-managers. How many different teams of 3 employees can be formed in which at least one member of the team is a manager and at least one member of the team is not a manager? (Two groups are considered different if at least one group member is different)

A. 84
B. 108
C. 135
D. 270
E. 990

I solved this task in this way:

(11!/3!*8!)-(3!+8!/3!5!)=103,
where (11!/3!*8!) is the total number of combinations and (3!+8!/3!5!) - undesirable output, e.g. only managers or non managers are in the teams
Could please someone to help me to point out, where exactly my logic comes wrong?
ShowHide Answer
Official Answer
B
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92921
Own Kudos [?]: 619088 [7]
Given Kudos: 81596
Send PM
CAT Tests Forum Quiz Mode
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   26 Apr 2012, 12:52
5
Kudos
2
Bookmarks
Expert Reply
Galiya wrote:
A certain company consists of 3 managers and 8 non-managers. How many different teams of 3 employees can be formed in which at least one member of the team is a manager and at least one member of the team is not a manager? (Two groups are considered different if at least one group member is different)

A. 84
B. 108
C. 135
D. 270
E. 990

I solved this task in this way:

(11!/3!*8!)-(3!+8!/3!5!)=103,
where (11!/3!*8!) is the total number of combinations and (3!+8!/3!5!) - undesirable output, e.g. only managers or non managers are in the teams
Could please someone to help me to point out, where exactly my logic comes wrong?


Direct approach:

Since there should be at least one manager and at least one non-manager among team of 3, then there should be either 1 manager and 2 non-managers OR 2 managers and 1 non-manager: \(C^1_3*C^2_8+C^2_3*C^1_8=84+24=108\).

Reverse approach:

Total # of teams of 3 possible is \(C^3_{11}=165\);
# of teams with only managers or only non-managers is: \(C^3_3+C^3_8=1+56=57\);

# of teams of 3 with at least one manager or at least one non-manager is: 165-57=108.

Answer: B.

Hope it's clear.
General Discussion
User avatar
L
Narenn
MBA Section Director
Joined: 22 Feb 2012
Affiliations: GMAT Club
Posts: 8701
Own Kudos [?]: 10013 [2]
Given Kudos: 4542
Test: Test
Send PM
GMAT ToolKit User
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   01 Jul 2013, 21:08
1
Kudos
1
Bookmarks
Expert Reply
Selection of a Manager from 3 Managers AND
Selection of 2 Non-Managers from 8 Non-Managers OR
Selection of 2 Managers from 3 Managers AND
Selection of a Non-Manager from 8 Non-Managers

AND = Multiplication, OR = Addition

3C1 * 8C2 + 3C2 * 8C1 = 3 * 28 + 3 * 8 = 24 + 84 = 108
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92921
Own Kudos [?]: 619088 [2]
Given Kudos: 81596
Send PM
CAT Tests Forum Quiz Mode
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   14 Aug 2013, 02:19
2
Kudos
Expert Reply
Bunuel wrote:
anoopv25 wrote:
first place can be filled with a manager in 3 ways
second place with a non manager in 8 ways
last place can be filled with either of them... i.e, 9 ways
so 3*8*9=216
where am i going wrong???


The following post might help: a-five-member-committee-is-to-be-formed-from-a-group-of-five-55410.html#p1248236


Check this too: a-committee-of-6-is-chosen-from-8-men-and-5-women-so-as-to-104859.html#p1046885
User avatar
B
galiya
Manager
Manager
Joined: 16 Jan 2011
Posts: 72
Own Kudos [?]: 715 [1]
Given Kudos: 15
Send PM
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   26 Apr 2012, 12:55
1
Kudos
i got it: instead of 3! i had to use 3!/3! for the combination of managers in a team
Bunuel! as always - thank you!
User avatar
B
yenh
Intern
Intern
Joined: 16 Mar 2014
Posts: 15
Own Kudos [?]: 48 [1]
Given Kudos: 123
GMAT Date: 08-18-2015
Send PM
GMAT ToolKit User Reviews Badge
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   06 May 2015, 20:23
1
Kudos
Hi Guys,
Here is my answer:
AT LEAST 1 manager and AT LEAST 1 non-manager. Don't be trapped by AT LEAST.
3 members with at least 1 manager that mean there can be ONE or TWO (cannot be three because there is AT LEAST 1 non-manager). Altogether, we have two cases:
(1) 1M + 2N = choose 1 of 3 + choose 2 of 8 = 3 x 8!/(2! x 6!) = 84
(2) 2M + 1N = choose 2 of 3 + choose 1 of 8 = 3!/(2! x 1!) x 8 = 24
So (1) + (2) = 108 -> B.
Have fun with Maths!
User avatar
PrepTap
Manager
Manager
Joined: 22 Apr 2015
Posts: 60
Own Kudos [?]: 409 [1]
Given Kudos: 6
Send PM
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   07 May 2015, 01:19
1
Kudos
reto wrote:
Bunuel wrote:
Galiya wrote:
A certain company consists of 3 managers and 8 non-managers. How many different teams of 3 employees can be formed in which at least one member of the team is a manager and at least one member of the team is not a manager? (Two groups are considered different if at least one group member is different)

A. 84
B. 108
C. 135
D. 270
E. 990

I solved this task in this way:

(11!/3!*8!)-(3!+8!/3!5!)=103,
where (11!/3!*8!) is the total number of combinations and (3!+8!/3!5!) - undesirable output, e.g. only managers or non managers are in the teams
Could please someone to help me to point out, where exactly my logic comes wrong?


Direct approach:

Since there should be at least one manager and at least one non-manager among team of 3, then there should be either 1 manager and 2 non-managers OR 2 managers and 1 non-manager: \(C^1_3*C^2_8+C^2_3*C^1_8=84+24=108\).

Reverse approach:

Total # of teams of 3 possible is \(C^3_{11}=165\);
# of teams with only managers or only non-managers is: \(C^3_3+C^3_8=1+56=57\);

# of teams of 3 with at least one manager or at least one non-manager is: 165-57=108.

Answer: B.

Hope it's clear.


Dear Bunuel

I do not understand the formulas here:

C13∗C28+C23∗C18=84+24=108. > How do you calculate 84 and 24 ... it must be with factorials but I do not get the way you write it. thank you


Hi reto

C13 is just another way of writing 3C1.

So the expression here can be converted to

3C1*8C2 + 3C2*8C1
= 3*(8*7/2) + 3*8
= 84 + 24
=108
avatar
raingary
Intern
Intern
Joined: 22 Jan 2012
Posts: 25
Own Kudos [?]: 76 [0]
Given Kudos: 0
Send PM
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   26 Apr 2012, 12:58
1M,2NM + 2M,1NM


3C1* 8C2 + 3C2 *8C1
3*28+3*8=108
avatar
casper123
Intern
Intern
Joined: 07 May 2012
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
Send PM
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   07 May 2012, 06:04
hi guys, first post :-D
i have a problem in understanding this question and why my logic is wrong, this is how i try to solve the problem:

i use the formula n!/(n-k)!*k!

so first n=3 and k=1 because we have 3 managers to choose from and can choose one
3!/(2!*1!)
second n=3 and k=2 because we have 3 managers to choose from and can choose two
3!/(2!*1!)
third n=8 and k=1 because we have 8 non-managers to choose from and can choose one
8!/(7!*1!)
fourth n=8 and k=2 because we have 8 non-managers to choose from and can choose two
8!/(6!*2!)
then i multiply all the parts= 3*3*8*28=2016
can someone please help me
thank you
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92921
Own Kudos [?]: 619088 [0]
Given Kudos: 81596
Send PM
CAT Tests Forum Quiz Mode
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   01 Jul 2013, 00:59
Expert Reply
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html
avatar
anoopv25
Intern
Intern
Joined: 07 Jun 2013
Posts: 3
Own Kudos [?]: 6 [0]
Given Kudos: 5
Location: India
Concentration: Entrepreneurship, Strategy
Schools: Oxford'17 (D) HEC '18 (A)
GMAT 1: 640 Q48 V31
GMAT 2: 710 Q35 V50
GPA: 3.2
WE:Business Development (Consulting)
Send PM
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   14 Aug 2013, 02:13
first place can be filled with a manager in 3 ways
second place with a non manager in 8 ways
last place can be filled with either of them... i.e, 9 ways
so 3*8*9=216
where am i going wrong???
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92921
Own Kudos [?]: 619088 [0]
Given Kudos: 81596
Send PM
CAT Tests Forum Quiz Mode
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   14 Aug 2013, 02:18
Expert Reply
anoopv25 wrote:
first place can be filled with a manager in 3 ways
second place with a non manager in 8 ways
last place can be filled with either of them... i.e, 9 ways
so 3*8*9=216
where am i going wrong???


The following post might help: a-five-member-committee-is-to-be-formed-from-a-group-of-five-55410.html#p1248236
User avatar
reto
Retired Moderator
Joined: 29 Apr 2015
Posts: 717
Own Kudos [?]: 4225 [0]
Given Kudos: 302
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE:Asset Management (Investment Banking)
Send PM
GMAT ToolKit User
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   06 May 2015, 13:37
Bunuel wrote:
Galiya wrote:
A certain company consists of 3 managers and 8 non-managers. How many different teams of 3 employees can be formed in which at least one member of the team is a manager and at least one member of the team is not a manager? (Two groups are considered different if at least one group member is different)

A. 84
B. 108
C. 135
D. 270
E. 990

I solved this task in this way:

(11!/3!*8!)-(3!+8!/3!5!)=103,
where (11!/3!*8!) is the total number of combinations and (3!+8!/3!5!) - undesirable output, e.g. only managers or non managers are in the teams
Could please someone to help me to point out, where exactly my logic comes wrong?


Direct approach:

Since there should be at least one manager and at least one non-manager among team of 3, then there should be either 1 manager and 2 non-managers OR 2 managers and 1 non-manager: \(C^1_3*C^2_8+C^2_3*C^1_8=84+24=108\).

Reverse approach:

Total # of teams of 3 possible is \(C^3_{11}=165\);
# of teams with only managers or only non-managers is: \(C^3_3+C^3_8=1+56=57\);

# of teams of 3 with at least one manager or at least one non-manager is: 165-57=108.

Answer: B.

Hope it's clear.


Dear Bunuel

I do not understand the formulas here:

C13∗C28+C23∗C18=84+24=108. > How do you calculate 84 and 24 ... it must be with factorials but I do not get the way you write it. thank you
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32679
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink] New post   10 Dec 2023, 21:35
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.
GMAT Club Bot
gmatclubot
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
10 Dec 2023, 21:35
Moderators:
Bunuel
Math Expert
92918 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions