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# A meeting has to be conducted with 5 managers. Find the

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Director
Joined: 28 Jul 2011
Posts: 512
Location: United States
GPA: 3.86
WE: Accounting (Commercial Banking)
A meeting has to be conducted with 5 managers. Find the [#permalink]

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02 May 2012, 22:53
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Difficulty:

45% (medium)

Question Stats:

67% (01:38) correct 33% (14:51) wrong based on 36 sessions

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A meeting has to be conducted with 5 managers. Find the number of ways in which the managers be selected from among 9 managers, if 2 managers will not attend the meeting together?

A. 35
B. 91
C. 120
D. 126
E. 150
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43901
Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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02 May 2012, 23:03
kotela wrote:
A meeting has to be conducted with 5 managers. Find the number of ways in which the managers be selected from among 9 managers, if 2 managers will not attend the meeting together?

A. 35
B. 91
C. 120
D. 126
E. 150

$$C^5_9-C^2_2*C^3_7=91$$, where:
$$C^5_9$$ is the total # of ways to choose 5 people out of 9 without any restriction;
$$C^2_2*C^3_7$$ is the # of ways to choose 2 managers who will not attend the meeting together and other 3 managers out 7 left (restriction).

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Intern
Joined: 01 May 2012
Posts: 3
Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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03 May 2012, 05:01
Could someone explain to me what $$C^5_9$$ is? I'm not familiar with anything that takes this form.

I got the right answer by doing 5! (5x4x3x2) and then knowing it had to be less than that because of the restriction but I don't understand what this solution means to do for getting the exact answer.
Math Expert
Joined: 02 Sep 2009
Posts: 43901
Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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03 May 2012, 11:19
DriftingAway wrote:
Could someone explain to me what $$C^5_9$$ is? I'm not familiar with anything that takes this form.

I got the right answer by doing 5! (5x4x3x2) and then knowing it had to be less than that because of the restriction but I don't understand what this solution means to do for getting the exact answer.

$$C^5_9=\frac{9!}{(9-5)!*5!}=\frac{9!}{4!*5!}$$ is the # of ways to choose 5 items out of 9 distinct items when the order of the selection does not matter.

Check this for more: math-combinatorics-87345.html

Hope it helps.
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Manager
Joined: 07 Feb 2008
Posts: 64
Location: United States
WE: Consulting (Computer Software)
Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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03 May 2012, 18:59
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i did this the other way around

we can either choose all 5 people from 7 manager who have no problems or choose 4 from the 7 and 1 from the 2 managers who have a problem sitting together

so 7C5 + (7C4 * 2C1)

this is 21 + 70

91
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Joined: 09 Sep 2013
Posts: 13759
Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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13 Apr 2017, 00:27
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Re: A meeting has to be conducted with 5 managers. Find the   [#permalink] 13 Apr 2017, 00:27
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