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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