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4 professors and 6 students are being considered for

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Director
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4 professors and 6 students are being considered for [#permalink] New post 06 Jun 2007, 19:49
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
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 [#permalink] New post 06 Jun 2007, 19:58
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, 12:37, edited 1 time in total.
Director
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Re: Probability [#permalink] New post 06 Jun 2007, 22:54
alimad wrote:
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
Manager
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 [#permalink] New post 06 Jun 2007, 23:08
at most means 0 or 1 professor. so again very simple
4C1*6C2+6C3=80
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 [#permalink] New post 07 Jun 2007, 13:52
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)

:-D
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 [#permalink] New post 10 Jun 2007, 14:16
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)

:-D


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.
Manager
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 [#permalink] New post 10 Jun 2007, 17:50
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?
CEO
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 [#permalink] New post 10 Jun 2007, 18:25
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.
GMAT Club Legend
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 [#permalink] New post 10 Jun 2007, 18:57
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.
  [#permalink] 10 Jun 2007, 18:57
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