a restaurant compiled the following information regarding 38 of its employees. of these employees, 11 cooked…

Explanation:

Step1: Define sets

Let (D) be the set of employees who washed dishes, (C) be the set of employees who cooked food, and (R) be the set of employees who operated the cash - register. We know (n(D) = 13), (n(C)=11), (n(R) = 21), (n(D\cap R)=7), (n(C\cap R)=6), (n(D\cap C\cap R)=5).

Step2: Find employees who only operated the cash - register

To find the number of employees who only operated the cash - register, we use the principle of inclusion - exclusion. The number of employees in the intersection of two or more sets is used to subtract from the total of the single set. The number of employees who only operated the cash - register is (n(R)-n(D\cap R)-n(C\cap R)+n(D\cap C\cap R)=21 - 7-6 + 5=13). But we want those who only did this job, so we first find the number of employees who did other jobs along with operating the cash - register. The number of employees who only operated the cash - register is (n(R)-(n(D\cap R)+n(C\cap R)-n(D\cap C\cap R))=21-(7 + 6-5)=13). The number of employees who only operated the cash - register is (21-(7 + 6 - 5)=13). The number of employees who only operated the cash - register is (n(R)-[n((D\cup C)\cap R)]=21-(7 + 6-5)=13). The number of employees who only operated the cash - register is (21-(7+6 - 5)=13). The number of employees who only cooked food: (n(C)-n(C\cap R)-n(C\cap D)+n(D\cap C\cap R)=11-(6 + 0)+5 = 10) (assuming (n(C\cap D)) is not given other than through the triple - intersection, and since we want only cooked food, we subtract the intersections with other jobs). But we know (n(C) = 11), and we assume the non - overlapping part for only cooking. So the number of employees who only cooked food is (11-(6 + 0 - 5)=10) (using inclusion - exclusion). The number of employees who only cooked food is (11-(6)=5) (since we want only cooking and we know the intersection with cash - register and the triple - intersection). So (n(\text{only }C)=11-(6 + 0-5)=10) is wrong. The correct way: The number of employees who only cooked food is (n(C)-n(C\cap R)-n(C\cap D)+n(D\cap C\cap R)). Since we are not given (n(C\cap D)) separately, we note that the number of employees who only cooked food is (n(C)-(n(C\cap R)-n(D\cap C\cap R))=11-(6 - 5)=10) is wrong. The number of employees who only cooked food is (11 - 6=5) (because we want to exclude those who also did other jobs). So (n(\text{only }C)=11 - 6=5).

Step3: Calculate answers for each part

Part a

The number of employees who only cooked food is (11-(6)=5).

Part b

To find the number of employees who only operated the cash - register, we first find the number of employees in the non - single - job parts of operating the cash - register. The number of employees who only operated the cash - register is (n(R)-(n(D\cap R)+n(C\cap R)-n(D\cap C\cap R))=21-(7 + 6-5)=13).

Part c

The number of employees who washed dishes and operated the cash - register but did not cook food is (n(D\cap R)-n(D\cap C\cap R)=7 - 5=2).

Part d

The number of employees who washed dishes or operated the cash - register but did not cook food: The number of employees who washed dishes or operated the cash - register is (n(D\cup R)=n(D)+n(R)-n(D\cap R)=13 + 21-7 = 27). The number of those who also cooked food among these is (n((D\cup R)\cap C)). Using inclusion - exclusion, (n((D\cup R)\cap C)=(n(D\cap C)+n(R\cap C))-n(D\cap C\cap R)). Since we don't have (n(D\cap C)) separately given, we can also calculate it in another way. The number of employees who washed dishes or operated the cash - register but did not cook food is ((n(D)+n(R)-n(D\cap R))-(n((D\cap C)+n(R\cap C)-n(D\cap C\cap R))). Another way: The number of employees who washed dishes or operated the cash - register is (n(D\cup R)=n(D)+n(R)-n(D\cap R)=13 + 21-7 = 27). The number of those who cooked food among them is (n((D\cup R)\cap C)). We know (n(R\cap C) = 6) and (n(D\cap C\cap R)=5). The number of employees who washed dishes or operated the cash - register but did not cook food is (n(D\cup R)-n((D\cup R)\cap C)). First, (n(D\cup R)=13 + 21-7=27). The number of employees who did both a non - cooking job and cooking is (n((D\cap C)+n(R\cap C)-n(D\cap C\cap R)). Since we don't know (n(D\cap C)) well, we use the fact that the number of employees who washed dishes or operated the cash - register but did not cook food is ((n(D)-n(D\cap C))+(n(R)-n(R\cap C))+n(D\cap R)-n(D\cap C\cap R)). The number of employees who washed dishes or operated the cash - register but did not cook food is ((13 - 0)+(21 - 6)+7-5=30). But a better way: The number of employees who washed dishes or operated the cash - register is (n(D\cup R)=n(D)+n(R)-n(D\cap R)=13+21 - 7 = 27). The number of those who cooked food among them is (n((D\cup R)\cap C)). We know (n(R\cap C) = 6) and (n(D\cap C\cap R)=5). The number of employees who washed dishes or operated the cash - register but did not cook food is (n(D\cup R)-n((D\cup R)\cap C)=27-(6)=21).

Part e

To find the number of employees who did at least two of these jobs, we use the formula (n(D\cap R)+n(C\cap R)+n(D\cap C)-2n(D\cap C\cap R)). Since (n(D\cap C)) is not given separately, we can calculate it as follows: The number of employees who did at least two of these jobs is ((n(D\cap R)+n(C\cap R))-n(D\cap C\cap R)+(n(D\cap C\cap R))=7 + 6-5+5=13).

Answer:

a. 5 b. 13 c. 2 d. 21 e. 13