A certain company consists of 3 managers and 8 non-managers. : Problem Solving (PS)

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

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Manager

Joined: 15 Jan 2011

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

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26 Apr 2012, 11:24

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45% (medium)

Question Stats:

67% (02:10) correct

33% (02:35) wrong

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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?

Spoiler: OA

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

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26 Apr 2012, 11:52

Galiya wrote:

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.
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Manager

Joined: 15 Jan 2011

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

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26 Apr 2012, 11:55

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!

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

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26 Apr 2012, 11:58

1M,2NM + 2M,1NM

3C1* 8C2 + 3C2 *8C1
3*28+3*8=108

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

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07 May 2012, 05:04

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

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

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30 Jun 2013, 23:59

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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01 Jul 2013, 20:08

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

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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???

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

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14 Aug 2013, 01:18

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

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14 Aug 2013, 01:19

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

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06 May 2015, 12:37

Bunuel wrote:

Galiya wrote:

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, 19:23

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!

Manager

Joined: 22 Apr 2015

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

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07 May 2015, 00:19

reto wrote:

Bunuel wrote:

Galiya wrote:

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

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

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