How Many Possible Committees with At Least Two Housekeepers Can a Resort Manager Form?

When managing a resort, it’s often necessary to form committees to discuss various issues, such as employment. A resort manager is currently tasked with selecting a committee of four people to discuss employment issues. The manager has a pool of candidates to choose from: eight housekeepers, three desk clerks, and five maintenance workers. The question is: how many possible committees can be formed where there are at least two housekeepers?

Breaking Down the Problem

Let's break down the problem into smaller, more manageable parts. We need to find all possible combinations where there are at least two housekeepers in the committee. We will consider three scenarios:

Case 1: Exactly Two Housekeepers in the Committee

When exactly two housekeepers are chosen, the remaining two positions are filled by either desk clerks or maintenance workers. Here are the steps to find the number of ways to form such a committee:

Pick 2 housekeepers out of 8: (binom{8}{2}) Pick 2 non-housekeepers (desk clerks or maintenance workers) out of 8: (binom{8}{2})

The total number of ways for this case is: (binom{8}{2} times binom{8}{2})

Case 2: Exactly Three Housekeepers in the Committee

When exactly three housekeepers are chosen, the remaining one position is filled by a non-housekeeper. Here are the steps to find the number of ways to form such a committee:

Pick 3 housekeepers out of 8: (binom{8}{3}) Pick 1 non-housekeeper out of 8: (binom{8}{1})

The total number of ways for this case is: (binom{8}{3} times binom{8}{1})

Case 3: Exactly Four Housekeepers in the Committee

When exactly four housekeepers are chosen, no other positions need to be filled. Here are the steps to find the number of ways to form such a committee:

Pick 4 housekeepers out of 8: (binom{8}{4})

The total number of ways for this case is: (binom{8}{4})

Calculating the Total Number of Possible Committees

Now, we combine the results from the three cases:

Case 1: (binom{8}{2} times binom{8}{2} 28 times 28 784) Case 2: (binom{8}{3} times binom{8}{1} 56 times 8 448) Case 3: (binom{8}{4} 70)

The total number of possible committees is the sum of the number of ways for each case:

Total number of committees 784 448 70 1302.

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