A committee, which consists of two mathematicians and three scientists, will be formed from four mathematicians and six scientists. How many possible committees can be formed if :

a. any of the mathematicians and scientists are included?
b. one particular scientist must be in the committee?
c. two particular mathematicians must be int he committee?
d. two particular scientists must not be int he committee?

a) number of (2math, 3 science() = C(4,2) x C(6,3) =120

b) one specific scientist must be in, so need 2 other scientist
= C(4,2) x C(6,2) = 90
c) 2 specific math guys must be in, but we need only 2 maths, so only need the 3 science
= C(6,3) = 20

d) 2 specific science guys are out, so need 3 from remaining 4 science guys
= C(4,2) x C(4,3) = 24

To solve each of the following questions, we can use combinations.

a. To find the number of possible committees when any of the mathematicians and scientists are included, we can choose any combination of mathematicians and scientists. Since there are four mathematicians and six scientists available, we can calculate the total number of configurations by summing the combinations for mathematicians and scientists separately.

The number of possible committees = (number of combinations for mathematicians) * (number of combinations for scientists)
= C(4,2) * C(6,3)
= 6 * 20
= 120

Therefore, there are 120 possible committees that can be formed.

b. To find the number of possible committees when one particular scientist must be in the committee, we need to choose the remaining scientists and the mathematicians from the remaining pool without restriction.

The number of possible committees = (number of combinations for remaining mathematicians) * (number of combinations for remaining scientists)
= C(3,2) * C(6,3)
= 3 * 20
= 60

Therefore, there are 60 possible committees that can be formed.

c. To find the number of possible committees when two particular mathematicians must be in the committee, we need to choose the remaining mathematician and the scientists without restriction.

The number of possible committees = (number of combinations for remaining mathematicians) * (number of combinations for scientists)
= C(2,1) * C(6,3)
= 2 * 20
= 40

Therefore, there are 40 possible committees that can be formed.

d. To find the number of possible committees when two particular scientists must not be in the committee, we need to exclude those two scientists and choose the committee members from the remaining mathematicians and scientists.

The number of possible committees = (number of combinations for mathematicians) * (number of combinations for remaining scientists)
= C(4,2) * C(4,3)
= 6 * 4
= 24

Therefore, there are 24 possible committees that can be formed.