From 5 statisticians and 6 economists, a committee consisting of 3 statisticians and 2 economists is to be formed. How many different committees can be formed if:
a. two particular statisticians must be in the committee?
b. one particular economist cannot be in the committee?
a. Well, if two particular statisticians must be in the committee, then they have no choice in the matter. So, we just need to choose the remaining statistician and the economists.
The remaining statistician can be chosen from the other 3 available, and the economists can be chosen from the remaining 6. Hence, the number of different committees that can be formed is 3C1 * 6C2 = 3 * (6! / (2!*(6-2)!)) = 3 * (6! / (2!*4!)) = 3 * (6 * 5 / (2 * 1)) = 3 * (30 / 2) = 3 * 15 = 45.
b. Now, if one particular economist cannot be in the committee, we need to consider two cases: the economist is chosen or the economist is not chosen.
Case 1: The economist is chosen. In this case, we need to choose 2 out of the remaining 5 economists, and 3 out of the 5 statisticians. Hence, the number of different committees formed with the particular economist is 5C2 * 5C3 = (5! / (2!*(3!))) * (5! / (3!*(2!))) = (5*4 / 2) * (5*4*3 / (3*2)) = 10 * 10 = 100.
Case 2: The economist is not chosen. In this case, we need to choose 2 economists from the remaining 5, and 3 statisticians from the available 5. Hence, the number of different committees formed without the particular economist is 5C2 * 5C3 = 100.
So, the total number of different committees that can be formed is 100 + 100 = 200.
a. To solve this problem, we need to consider the number of ways to select the required number of statisticians and economists separately.
There are 5 statisticians in total, and we need to select 3 of them. We can calculate this using the combination formula:
C(n, r) = n! / (r! * (n-r)!)
C(5, 3) = 5! / (3! * (5-3)!) = 5! / (3! * 2!)
= (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (2 * 1))
= (5 * 4) / (2 * 1)
= 10 / 2
= 5
So, there are 5 ways to select 3 statisticians from the 5 available.
Next, there are 6 economists in total, and we need to select 2 of them. Using the same combination formula:
C(6, 2) = 6! / (2! * (6-2)!)
= 6! / (2! * 4!)
= (6 * 5 * 4 * 3 * 2 * 1) / ((2 * 1) * (4 * 3 * 2 * 1))
= (6 * 5) / (2 * 1)
= 30 / 2
= 15
So, there are 15 ways to select 2 economists from the 6 available.
To calculate the total number of committees, we multiply the number of ways to select statisticians with the number of ways to select economists:
Total number of committees = Number of ways to select statisticians * Number of ways to select economists
= 5 * 15
= 75
Therefore, there are 75 different committees that can be formed if two particular statisticians must be in the committee.
b. To solve this problem, we will subtract the number of committees where the particular economist is included from the total number of committees.
First, let's calculate the total number of committees without any restrictions:
Total number of committees = Number of ways to select 3 statisticians * Number of ways to select 2 economists
= 5 * 15
= 75
Next, we need to calculate the number of committees that include the particular economist. Since exactly 2 economists are needed for the committee, the particular economist must be included in each committee. So, we just need to select the remaining 1 economist from the remaining economists:
Number of committees including the particular economist = Number of ways to select 1 economist from the remaining 5 economists
= C(5, 1)
= 5! / (1! * (5-1)!)
= 5! / (1! * 4!)
= (5 * 4 * 3 * 2 * 1) / ((1 * 1) * (4 * 3 * 2 * 1))
= 5
Therefore, to find the number of committees where one particular economist cannot be in the committee, we subtract the number of committees including that economist from the total number of committees:
Number of committees without the particular economist = Total number of committees - Number of committees including the particular economist
= 75 - 5
= 70
Therefore, there are 70 different committees that can be formed if one particular economist cannot be in the committee.
To solve this problem, we can use combinations.
a. To find the number of different committees that can be formed when two particular statisticians must be in the committee, we need to consider two cases:
1) The two particular statisticians are already chosen, so we need to choose the remaining 1 statistician from the remaining 5 statisticians.
2) Both the particular statisticians are not chosen, so we need to choose 3 statisticians from the remaining 5 statisticians.
For case 1: the number of ways to choose 2 particular statisticians is 1 (since they are selected already) and the number of ways to choose the remaining 1 statistician from the remaining 5 is given by C(5, 1) = 5.
For case 2: the number of ways to choose 2 particular statisticians is given by C(5, 2) = 10 and the number of ways to choose the remaining 1 statistician from the remaining 5 is given by C(5, 1) = 5.
So, the total number of different committees that can be formed when two particular statisticians must be in the committee is 1 * 5 + 10 * 5 = 1 * 5 + 10 * 5 = 55.
b. To find the number of different committees that can be formed when one particular economist cannot be in the committee, we need to consider two cases:
1) The particular economist is not chosen, so we need to choose 2 economists from the remaining 5 economists.
2) The particular economist is chosen, so we need to choose 2 economists from the remaining 5 economists instead of 6.
For case 1: the number of ways to choose 2 economists from the remaining 5 is given by C(5, 2) = 10.
For case 2: the number of ways to choose 2 economists from the remaining 5 (considering only 5 economists) is given by C(5, 2) = 10.
So, the total number of different committees that can be formed when one particular economist cannot be in the committee is 10 + 10 = 20.
Therefore, the answers are:
a. 55 different committees can be formed if two particular statisticians must be in the committee.
b. 20 different committees can be formed if one particular economist cannot be in the committee.
a 200
b 565