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 answer the first part of the question, we need to calculate the number of different arrangements possible for selecting a committee of twelve from a group of nineteen people. This can be done using the concept of combinations.

The total number of combinations, denoted as C(n, r), where n represents the total number of items, and r represents the number of items selected, can be calculated using the formula:

C(n, r) = n! / (r!(n-r)!)

In this case, we have n = 19 (total number of club members) and r = 12 (number of committee members to be selected). Substituting these values into the formula, we can calculate the total number of arrangements as follows:

C(19, 12) = 19! / (12!(19-12)!)
= 19! / (12! * 7!)
= (19 * 18 * 17 * 16 * 15 * 14 * 13) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
= 19 * 3 * 17 * 2 * 13
= 34,068

Therefore, there are 34,068 different arrangements possible when selecting a committee of twelve from a group of nineteen people.

Moving on to the second part of the question, we need to determine how many of these committees will contain Bill, Mary, and Joe at the same time. Since Bill, Mary, and Joe must be included in the committee, we now need to select the remaining nine members from the remaining sixteen people.

Using the same formula as before, but with n = 16 and r = 9, we can calculate the number of committees that include Bill, Mary, and Joe:

C(16, 9) = 16! / (9!(16-9)!)
= 16! / (9! * 7!)
= (16 * 15 * 14 * 13 * 12 * 11 * 10) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
= 32,768

Therefore, there are 32,768 different committees that will contain Bill, Mary, and Joe at the same time.