Problem 27 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(A, B\), and \(C\) be subsets of a universal set \(U\) and suppose \(n(U)=100, n(A)=28, n(B)=30\) \(n(C)=34, n(A \cap B)=8, n(A \cap C)=10, n(B \cap C)=15\) and \(n(A \cap B \cap C)=5\). Compute: a. \(n\left(A^{c} \cap B^{c} \cap C^{c}\right)\) b. \(n\left[A^{c} \cap(B \cup C)\right]\)

Step-by-Step Solution

Answer

a. \(n\left(A^{c} \cap B^{c} \cap C^{c}\right) = 36\) b. \(n\left[A^{c} \cap(B \cup C)\right] = 41\)

Step 1: a. Compute the number of elements in \(A^{c} \cap B^{c} \cap C^{c}\).

To compute the number of elements in \(A^{c} \cap B^{c} \cap C^{c}\), we will use the fact that: \(n(A^{c} \cap B^{c} \cap C^{c}) = n(U) - n(A \cup B \cup C)\) Now, we can calculate the number of elements in the union of A, B, and C using the Inclusion-Exclusion Principle: \(n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)\)

Step 2: Calculate \(n(A \cup B \cup C)\).

Substituting the given values, we have: \(n(A \cup B \cup C) = 28 + 30 + 34 - 8 - 10 - 15 + 5 = 64\)

Step 3: Calculate \(n\left(A^{c} \cap B^{c} \cap C^{c}\right)\).

Now, we can compute the number of elements in the intersection of the complements of A, B, and C: \(n\left(A^{c} \cap B^{c} \cap C^{c}\right) = n(U) - n(A \cup B \cup C) = 100 - 64 = 36\) Thus, \(n\left(A^{c} \cap B^{c} \cap C^{c}\right) = 36\).

Step 4: b. Compute the number of elements in \(A^{c} \cap (B \cup C)\).

We can rewrite the given expression as: \(n\left(A^{c} \cap (B \cup C)\right) = n\left((U \setminus A) \cap (B \cup C)\right)\) by the definition of the complement of a set. We then apply De Morgan's Laws: \(n\left((U \setminus A) \cap (B \cup C)\right) = n\left(U \cap (B \cup C) \setminus A \cap (B \cup C)\right)\) Using the fact that \(U \cap (B \cup C) = (B \cup C)\), we have: \(n\left(A^{c} \cap (B \cup C)\right) = n(B \cup C) - n(A \cap (B \cup C))\)

Step 5: Calculate \(n(B \cup C)\).

We will now calculate the number of elements in \(B \cup C\): \(n(B \cup C) = n(B) + n(C) - n(B \cap C) = 30 + 34 - 15 = 49\)

Step 6: Calculate \(n\left(A \cap (B \cup C)\right)\).

We will now calculate the number of elements in \(A \cap (B \cup C)\): \(n\left(A \cap (B \cup C)\right) = n\left(A \cap (B \cap C^{c}) \cup (B^{c} \cap C)\right)\) Recall that: \(n\left(A \cap (B \cap C^{c}) \cup (B^{c} \cap C)\right) = n(A \cap B \cap C^{c}) + n(A \cap B^{c} \cap C) - n(A \cap B \cap C)\) Substituting the given values, we have: \(n\left(A \cap (B \cup C)\right) = 8 - 5 + 10 - 5 = 8\)

Step 7: Calculate \(n\left[A^{c} \cap(B \cup C)\right]\).

Now, we can compute the number of elements in \(A^{c} \cap (B \cup C)\) using the formula we derived earlier: \(n\left[A^{c} \cap(B \cup C)\right] = n(B \cup C) - n(A \cap (B \cup C)) = 49 - 8 = 41\) Thus, \(n\left[A^{c} \cap(B \cup C)\right] = 41\).

Key Concepts

Inclusion-Exclusion Principle

The Inclusion-Exclusion Principle is a fundamental concept in set theory used to calculate the size of the union of multiple sets. When you have three sets, say \(A\), \(B\), and \(C\), the principle helps you avoid counting elements that belong to more than one set multiple times. To find \(n(A \cup B \cup C)\), you can apply the formula:

• Start by adding the sizes of the individual sets: \(n(A) + n(B) + n(C)\).
• Then, subtract the sizes of the pairwise intersections to remove the double-counted elements: \(- n(A \cap B) - n(A \cap C) - n(B \cap C)\).
• Finally, add back the size of the intersection of all three sets to fix any elements removed too many times: \(+ n(A \cap B \cap C)\).

This comprehensive approach ensures you count each element exactly once for the union. In our specific problem, this principle allowed us to find that \(n(A \cup B \cup C) = 64\).

Universal Set

A universal set, denoted often as \(U\), is the set that contains all objects under consideration. It serves as the "reference" set in discussions about other sets and their relationships. All other sets, like \(A\), \(B\), and \(C\), are considered subsets of this universal set.

• In our problem, the universal set \(U\) had 100 elements, i.e., \(n(U) = 100\).
• This universal set allows us to determine the complement of any subset easily.
• Anything not included in a subset is its complement with respect to \(U\).

Knowing the size of \(U\) helps us compute the number of elements in a complement, such as finding \(n(A^c) = n(U) - n(A)\). This helps solve complex set problems like determining the size of \(A^c \cap B^c \cap C^c\).

De Morgan's Laws

De Morgan's Laws are two transformation rules that relate the complements of unions and intersections of sets. These laws are very useful for simplifying expressions involving complements in set theory.

• The first law states: \((A \cap B)^c = A^c \cup B^c\).
• The second law states: \((A \cup B)^c = A^c \cap B^c\).

They can be extended to more than two sets. In our exercise, these laws were implicit in simplifying the expression \(A^c \cap (B \cup C)\). By applying De Morgan's Laws, complex set relations become easier to handle and understand, particularly when switching between complements and their direct counterparts.

Complement of a Set

The complement of a set refers to all the elements that are in the universal set \(U\) but not in a given subset. It's a crucial concept because it offers a different perspective of the set – what it doesn't include.

• If \(A\) is a subset of \(U\), its complement \(A^c\) consists of all the elements in \(U\) that are not in \(A\).
• Mathematically, \(n(A^c) = n(U) - n(A)\).

Consider the problem at hand; calculating \(n(A^c \cap B^c \cap C^c)\) involved recognizing that this intersects the complements of all three sets. The calculation used the relation \(n(A^c \cap B^c \cap C^c) = n(U) - n(A \cup B \cup C)\), which is a direct application of the complement's definition. Complements help find out how much is "left out" when specific elements are accounted for by a set.