Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 16 Jan 2011
Posts: 103

A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
26 Apr 2012, 12:24
1
This post received KUDOS
2
This post was BOOKMARKED
Question Stats:
70% (02:37) correct
30% (02:24) wrong based on 163 sessions
HideShow timer Statistics
A certain company consists of 3 managers and 8 nonmanagers. 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?
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
26 Apr 2012, 12:52
2
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
Galiya wrote: A certain company consists of 3 managers and 8 nonmanagers. 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 nonmanager among team of 3, then there should be either 1 manager and 2 nonmanagers OR 2 managers and 1 nonmanager: \(C^1_3*C^2_8+C^2_3*C^1_8=84+24=108\). Reverse approach:Total # of teams of 3 possible is \(C^3_{11}=165\); # of teams with only managers or only nonmanagers is: \(C^3_3+C^3_8=1+56=57\); # of teams of 3 with at least one manager or at least one nonmanager is: 16557=108. Answer: B. Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 16 Jan 2011
Posts: 103

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
26 Apr 2012, 12:55
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!



Intern
Joined: 22 Jan 2012
Posts: 35

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
26 Apr 2012, 12:58
1M,2NM + 2M,1NM
3C1* 8C2 + 3C2 *8C1 3*28+3*8=108



Intern
Joined: 07 May 2012
Posts: 1

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
07 May 2012, 06:04
hi guys, first post 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!/(nk)!*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 nonmanagers to choose from and can choose one 8!/(7!*1!) fourth n=8 and k=2 because we have 8 nonmanagers 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



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
01 Jul 2013, 00:59



MBA Section Director
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 4390
Location: India
City: Pune
GPA: 3.4
WE: Business Development (Manufacturing)

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
01 Jul 2013, 21:08
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
Selection of a Manager from 3 Managers ANDSelection of 2 NonManagers from 8 NonManagers ORSelection of 2 Managers from 3 Managers ANDSelection of a NonManager from 8 NonManagers AND = Multiplication, OR = Addition 3C1 * 8C2 + 3C2 * 8C1 = 3 * 28 + 3 * 8 = 24 + 84 = 108
_________________
Every Wednesday: Meet MBA Experts in Chat Room and Ask Your MostPressing MBA Admission Questions to them in a Live Chat.
Must Read Forum Topics Before You Kick Off Your MBA Application
New GMAT Club Decision Tracker  Real Time Decision Updates



Current Student
Joined: 07 Jun 2013
Posts: 3
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 640 Q48 V31 GMAT 2: 710 Q35 V50
GPA: 3.2
WE: Business Development (Consulting)

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
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???



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
14 Aug 2013, 02:18



Math Expert
Joined: 02 Sep 2009
Posts: 39589

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
14 Aug 2013, 02:19



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15916

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
06 May 2015, 06:42
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 Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Verbal Forum Moderator
Joined: 29 Apr 2015
Posts: 897
Location: Switzerland
Concentration: Economics, Finance
WE: Asset Management (Investment Banking)

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
06 May 2015, 13:37
Bunuel wrote: Galiya wrote: A certain company consists of 3 managers and 8 nonmanagers. 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 nonmanager among team of 3, then there should be either 1 manager and 2 nonmanagers OR 2 managers and 1 nonmanager: \(C^1_3*C^2_8+C^2_3*C^1_8=84+24=108\). Reverse approach:Total # of teams of 3 possible is \(C^3_{11}=165\); # of teams with only managers or only nonmanagers is: \(C^3_3+C^3_8=1+56=57\); # of teams of 3 with at least one manager or at least one nonmanager is: 16557=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
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.



Intern
Joined: 16 Mar 2014
Posts: 16
GMAT Date: 08182015

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
06 May 2015, 20:23
1
This post received KUDOS
Hi Guys, Here is my answer: AT LEAST 1 manager and AT LEAST 1 nonmanager. 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 nonmanager). 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!



Manager
Joined: 22 Apr 2015
Posts: 66

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
07 May 2015, 01:19
1
This post received KUDOS
reto wrote: Bunuel wrote: Galiya wrote: A certain company consists of 3 managers and 8 nonmanagers. 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 nonmanager among team of 3, then there should be either 1 manager and 2 nonmanagers OR 2 managers and 1 nonmanager: \(C^1_3*C^2_8+C^2_3*C^1_8=84+24=108\). Reverse approach:Total # of teams of 3 possible is \(C^3_{11}=165\); # of teams with only managers or only nonmanagers is: \(C^3_3+C^3_8=1+56=57\); # of teams of 3 with at least one manager or at least one nonmanager is: 16557=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



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15916

Re: A certain company consists of 3 managers and 8 nonmanagers. [#permalink]
Show Tags
14 May 2016, 03:35
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 Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: A certain company consists of 3 managers and 8 nonmanagers.
[#permalink]
14 May 2016, 03:35







