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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] New post   02 May 2012, 23:53
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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
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Re: A meeting has to be conducted with 5 managers. Find the [#permalink] New post   03 May 2012, 00:03
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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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Re: A meeting has to be conducted with 5 managers. Find the [#permalink] 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] New post   03 May 2012, 12:19
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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] 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
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Re: A meeting has to be conducted with 5 managers. Find the [#permalink] New post   01 Oct 2020, 21:46
(Total No. of Ways to choose ANY 5 with NO Restriction) - (No. of Ways to Choose a Team where 2 MEN are ALWAYS Selected) = No. of Ways to Choose Team in which 2 Men are NEVER on the Same Team

Place the 2 Men on the Team for each Possible Combination (only have to Select 3 now) and Reduce the Overall Pool Available by -2 also.

"9 choose 5" - "7 choose 3" = 9! / (5!4!) - 7! / (4!3!) = 126 - 35 = 91 Ways

-B-
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Re: A meeting has to be conducted with 5 managers. Find the [#permalink] New post   02 Oct 2020, 03:34
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Total managers: 9

To be selected: 5

Number of ways: \(^9{C_5}\) = 126

If '2' managers always attend meeting together, then we will require only 3 managers to select out of '7': \(^7{C_3}\) = 35

Number of ways in which '2' managers will not attend the meeting together: 126 - 35 = 91

Answer B
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Re: A meeting has to be conducted with 5 managers. Find the [#permalink] New post   06 Oct 2023, 03:23
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