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A certain company consists of 3 managers and 8 non-managers. [#permalink]

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26 Apr 2012, 12:24

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72% (02:34) correct
28% (02:30) wrong based on 135 sessions

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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]

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26 Apr 2012, 12:52

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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]

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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!/(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]

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01 Jul 2013, 21:08

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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]

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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???

Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]

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14 Aug 2013, 02:18

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]

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14 Aug 2013, 02:19

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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]

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06 May 2015, 06:42

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Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]

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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 _________________

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Re: A certain company consists of 3 managers and 8 non-managers. [#permalink]

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06 May 2015, 20:23

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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]

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07 May 2015, 01:19

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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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