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A certain company consists of 3 managers and 8 non-managers. [#permalink]
26 Apr 2012, 11:24
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Question Stats:
73% (02:35) correct
27% (02:25) wrong based on 131 sessions
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?
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
26 Apr 2012, 11:52
1
This post received KUDOS
Expert's post
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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.
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
07 May 2012, 05:04
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!/(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
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
01 Jul 2013, 20:08
1
This post received KUDOS
Expert's post
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
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
14 Aug 2013, 01:13
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???
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
14 Aug 2013, 01:18
Expert's post
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???
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
14 Aug 2013, 01:19
1
This post received KUDOS
Expert's post
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???
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
06 May 2015, 05:42
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Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
06 May 2015, 12:37
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 _________________
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.
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
06 May 2015, 19:23
1
This post received 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!
Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]
07 May 2015, 00:19
1
This post received 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
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