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Re: 4 professors and 6 students are being considered for [#permalink]
at most means 0 or 1 professor. so again very simple
4C1*6C2+6C3=80
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Re: 4 professors and 6 students are being considered for [#permalink]
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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Re: 4 professors and 6 students are being considered for [#permalink]
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.
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Re: 4 professors and 6 students are being considered for [#permalink]
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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Re: 4 professors and 6 students are being considered for [#permalink]
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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Re: 4 professors and 6 students are being considered for [#permalink]
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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