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# 4 professors and 6 students are being considered for

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Director
Joined: 10 Feb 2006
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4 professors and 6 students are being considered for [#permalink]

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06 Jun 2007, 20:49
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committee be formed if it has to include at most 1 professor ?
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Manager
Joined: 28 Aug 2006
Posts: 159

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06 Jun 2007, 20:58
At most 1 professor means 1 or no professors

No professors, just students 6C3 ways
1 prof and 2 students 4C1.6C2

Total ways =6C3+4C1.6C2

Sorry Probablity is

=80/120 =>3/4

Last edited by vijay2001 on 07 Jun 2007, 13:37, edited 1 time in total.

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Director
Joined: 03 Sep 2006
Posts: 865

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06 Jun 2007, 23:54
4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committee be formed if it has to include at most 1 professor ?

P :4 S:4

Total ways to select 3 people = 10C3 = 120

No of fav. outcomes = 4C3 + 3C1*4C1 = 16

therefore P = 16/120 = 2/15

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Manager
Joined: 17 Oct 2006
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07 Jun 2007, 00:08
at most means 0 or 1 professor. so again very simple
4C1*6C2+6C3=80

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VP
Joined: 08 Jun 2005
Posts: 1144

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07 Jun 2007, 14:52
shahrukh wrote:
at most means 0 or 1 professor. so again very simple
4C1*6C2+6C3=80

Perfect !

4C1*6C2+4C0*6C3=80

(choosing one professor out of four * choosing students as the rest) + (choosing zero professors out of four * choosing students as the rest)

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Intern
Joined: 09 Jun 2007
Posts: 2

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10 Jun 2007, 15:16
KillerSquirrel wrote:
shahrukh wrote:
at most means 0 or 1 professor. so again very simple
4C1*6C2+6C3=80

Perfect !

4C1*6C2+4C0*6C3=80

(choosing one professor out of four * choosing students as the rest) + (choosing zero professors out of four * choosing students as the rest)

any way you could please explain this a little more in depth, pretending that you were explaining to someone who has never done a permutation/combination problem before? I'd greatly appreciate this.

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Manager
Joined: 14 Mar 2007
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10 Jun 2007, 18:50
Quote:
4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committee be formed if it has to include at most 1 professor ?

PLEASE TELL ME WHY CAN´T i FIGURE THIS PROBLEM OUT LIKE BELOW: i SAW 80 IN OTHER POSTS.....

6*( 6*5*4)+ (6*5*4)
= 840 WAYS????
_ _ _
6 5 4 1 PROFESSOR
6 4 5
5 6 4
5 4 6
4 6 5
4 5 6

6 5 4 0 PROFESSOR

WHAT IS THE OA?

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CEO
Joined: 17 May 2007
Posts: 2947

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10 Jun 2007, 19:25
For those who want a detailed explanation :

How many ways can 6 students form a committee of 3 ?
6C3 --1
How many ways can 4 professors form 1 member of the committee ?
4C1 --2
How many ways can 6 students form 2 members of the committee ?
6C2 --3
Total ways is
1 + 2*3
6C3 + 6C2*4C1 = 80

Hope this helps. In terms of how to solve 6C3 etc., you are better of reading a basic combinatorics book.

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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5034

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Location: Singapore

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10 Jun 2007, 19:57
Each professor can pair up with 6C2 students = 15 combinations
4 professors will give 15*4 =60 combinations.
Also, we have committees formed entirely from student body = 6C3 = 20 such combinations.

Total = 80 groups.

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10 Jun 2007, 19:57
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