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?
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 ??
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
its 9 mathematician instead of 5. Sorry again.
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.
Thanks much. I will do just that.
To solve these types of combinatorial problems, we can use the concept of combinations.
A combination is a way to select items from a larger set without considering their order. The formula to find the number of combinations is:
nCr = n! / r!(n-r)!
Where n is the total number of items, r is the number of items to be selected, and ! denotes factorial (the product of all positive integers from 1 to n).
Let's solve each part of the problem using this formula:
a. All scientists:
Since there are 5 scientists, we need to select all of them. So, the number of committees that can be formed is the combination of selecting 5 scientists from 5.
nCr = 5! / 5!(5-5)! = 5! / 5!0! = 1
Therefore, there is only 1 committee that can be formed when all members are scientists.
b. Exactly 3 mathematicians:
There are 9 mathematicians, and we need to select exactly 3 of them. The number of committees that can be formed is the combination of selecting 3 mathematicians from 9.
nCr = 9! / 3!(9-3)! = 9! / 3!6! = (9 * 8 * 7) / (3 * 2 * 1) = 84
Therefore, 84 committees can be formed with exactly 3 mathematicians.
c. 2 scientists are not allowed:
In this case, we have 5 scientists, and we need to select a committee from the remaining 3 scientists out of the total 5. The number of committees that can be formed is the combination of selecting 3 scientists from 5.
nCr = 5! / 3!(5-3)! = 5! / 3!2! = (5 * 4) / (2 * 1) = 10
Therefore, 10 committees can be formed when 2 particular scientists are not allowed.
d. 1 scientist and 1 mathematician are not allowed:
In this case, we have to consider the exclusion of 1 scientist and 1 mathematician. So, we need to select a committee from the remaining 4 scientists and 8 mathematicians out of the total 5 scientists and 9 mathematicians.
nCr = (4 + 8)! / 3!(4+8-3)! = 12! / 3!9! = (12 * 11 * 10) / (3 * 2 * 1) = 220
Therefore, 220 committees can be formed when 1 scientist and 1 mathematician are not allowed.
e. 2 mathematicians and 1 scientist cannot be in the committee:
In this case, we have to exclude 2 mathematicians and 1 scientist from the total 5 scientists and 9 mathematicians. We need to select a committee from the remaining 3 scientists and 7 mathematicians.
nCr = (3 + 7)! / 3!(3+7-3)! = 10! / 3!7! = (10 * 9 * 8) / (3 * 2 * 1) = 120
Therefore, 120 committees can be formed when 2 mathematicians and 1 scientist cannot be in the committee.