Bill,mary,and Joe belong to a club of nineteen people. A committee of twelve is to be selected at random from the membership. How many different arrangements are possible? How many of these committees will contain Bill,Mary, and Joe at the same time?
To find the number of different arrangements possible, we need to use combinations. The formula for combinations is given by:
C(n, r) = n! / (r! * (n-r)!)
Where n is the total number of individuals (19) and r is the number of people to be selected for the committee (12).
To calculate the number of different arrangements possible, we substitute the values in the formula:
C(19, 12) = 19! / (12! * (19-12)!)
= 19! / (12! * 7!)
Now, let's calculate the value using this formula:
19! = 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12!
Therefore, the calculation becomes:
C(19, 12) = (19 * 18 * 17 * 16 * 15 * 14 * 13 * 12!) / (12! * 7!)
The 12! terms cancel out, leaving us with:
C(19, 12) = 19 * 18 * 17 * 16 * 15 * 14 * 13 / 7!
To find the factorial of 7, we calculate:
7! = 7 * 6 * 5 * 4 * 3 * 2 * 1
Now we substitute the value in the formula:
C(19, 12) = (19 * 18 * 17 * 16 * 15 * 14 * 13) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
Calculating this gives us the number of different arrangements possible, which is:
C(19, 12) = 3,535,316 arrangements
Now, to find the number of committees that will contain Bill, Mary, and Joe at the same time, we need to calculate the number of committees that can be formed from the remaining members. Since Bill, Mary, and Joe have already been selected, we now need to choose 9 members from the remaining 16.
Using the same combination formula, we can calculate:
C(16, 9) = 16! / (9! * (16-9)!)
Simplifying this equation, we obtain:
C(16, 9) = (16 * 15 * 14 * 13 * 12 * 11 * 10 * 9!) / (9! * 7!)
The 9! terms cancel out, leaving us with:
C(16, 9) = (16 * 15 * 14 * 13 * 12 * 11 * 10) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
Calculating this gives us the number of committees that will contain Bill, Mary, and Joe, which is:
C(16, 9) = 7,329 committees.
Therefore, there are 3,535,316 different arrangements possible from the 19 members, and out of these arrangements, 7,329 committees will contain Bill, Mary, and Joe at the same time.