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

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

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New post 02 May 2012, 23:53
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Question Stats:

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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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Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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New post 03 May 2012, 00:03
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).

Answer: B.
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New post 03 May 2012, 06:01
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.
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Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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New post 03 May 2012, 12:19
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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Re: A meeting has to be conducted with 5 managers. Find the [#permalink]

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New post 03 May 2012, 19:59
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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
Re: A meeting has to be conducted with 5 managers. Find the   [#permalink] 03 May 2012, 19:59
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