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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
[Reveal] Spoiler: OA

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Re: A meeting has to be conducted with 5 managers. Find the [#permalink] 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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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
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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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