Asked by Therese
                Given 5 members, 5 scientists and 9 mathematicians.  How many committees can be formed if ..... 
a. all are scientists?
b. exactly 3 mathematicians?
c. 2 scientists are not allowed to be in the committee?
d. 1 of the scientists and 1 of the mathematicians are not allowed to be in the committee?
e. 2 mathematicians and 1 of the scientists cannot be in the committee?
            
        a. all are scientists?
b. exactly 3 mathematicians?
c. 2 scientists are not allowed to be in the committee?
d. 1 of the scientists and 1 of the mathematicians are not allowed to be in the committee?
e. 2 mathematicians and 1 of the scientists cannot be in the committee?
Answers
                    Answered by
            Reiny
            
    Before I answer, clear up the following:
We have 14 people, 5 of them scientists and 9 mathematicans, and we want to form a committee of 5 ??
    
We have 14 people, 5 of them scientists and 9 mathematicans, and we want to form a committee of 5 ??
                    Answered by
            Therese
            
    ooops sorry.  This is a more clear question.
How many different committee with 5 members can be formed if you have to choose from 5 scientists, 5 mathematicians?
a. all are scientists?
b. exactly 3 mathematicians?
c. 2 scientists are not allowed to be in the committee?
d. 1 of the scientists and 1 of the mathematicians are not allowed to be in the committee?
e. 2 mathematicians and 1 of the scientists cannot be in the committee
    
How many different committee with 5 members can be formed if you have to choose from 5 scientists, 5 mathematicians?
a. all are scientists?
b. exactly 3 mathematicians?
c. 2 scientists are not allowed to be in the committee?
d. 1 of the scientists and 1 of the mathematicians are not allowed to be in the committee?
e. 2 mathematicians and 1 of the scientists cannot be in the committee
                    Answered by
            Therese
            
    its 9 mathematician instead of 5.  Sorry again.
    
                    Answered by
            Reiny
            
    That's better, BUT did you mean to change the number of mathematicians to 5 from 9 ????
I will assume you meant 9 maths
a) since you need all 5 scientists in your committee of 5,
number of ways = C(5,5) = 1
b) so you want 3 of the 5 scientists and 2 of the maths
number of ways = C(5,3) x C(9,2) = 10(36) = 360
c) so we just eliminate 2 of the people
seems to be no other restriction, simply choose 5 from the remaining 12
= C(12,5) = 792
d) to me that is the same as c)
e) 3 specified people are out, so leaves 11 to choose from
C(11,5) = 462
If you meant to change the number of math guys from 9 to 5, just make the necessary changes in the above steps.
    
I will assume you meant 9 maths
a) since you need all 5 scientists in your committee of 5,
number of ways = C(5,5) = 1
b) so you want 3 of the 5 scientists and 2 of the maths
number of ways = C(5,3) x C(9,2) = 10(36) = 360
c) so we just eliminate 2 of the people
seems to be no other restriction, simply choose 5 from the remaining 12
= C(12,5) = 792
d) to me that is the same as c)
e) 3 specified people are out, so leaves 11 to choose from
C(11,5) = 462
If you meant to change the number of math guys from 9 to 5, just make the necessary changes in the above steps.
                    Answered by
            Therese
            
    Thanks much.  I will do just that.  
    
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