Solving Set Theory Problem: Finding n

This problem requires us to use the principle of double counting to find the value of \(n\), which represents the number of sets in the collection \(B_j\). We are given details about two collections of sets, \(A_i\) and \(B_j\), their sizes, and how elements are distributed among them within a universal set \(S\).

Understanding the Given Information

• We have 30 sets, named \(A_1, A_2, ..., A_{30}\).
• Each of these 30 sets contains exactly 5 elements.
• We also have \(n\) sets, named \(B_1, B_2, ..., B_n\).
• Each of these \(n\) sets contains exactly 3 elements.
• The union of all sets \(A_i\) is the same as the union of all sets \(B_j\). This combined union forms the set \(S\). This can be written as:

  \(\displaystyle\bigcup^{30}_{i = 1} A_i = \displaystyle\bigcup^n_{j = 1} B_j = S\)

• A crucial condition is that every single element within the set \(S\) is a member of exactly 10 different sets from the collection \(A_i\).
• Similarly, every element within the set \(S\) is also a member of exactly 9 different sets from the collection \(B_j\).

Applying the Principle of Double Counting

The Principle of Double Counting is a useful technique in combinatorics. It involves counting the same quantity in two different ways to establish an equality. We will use it here to find the size of the set \(S\) and then determine \(n\).

The core idea is:

Total count = Sum of sizes of individual sets

Total count = Sum over all elements (Number of sets the element belongs to)

Calculating the Size of Set S

Let's first find the total number of elements in the set \(S\) using the information about the \(A_i\) sets.

• The sum of the number of elements across all 30 sets \(A_i\) is:

  \(\sum_{i=1}^{30} |A_i| = 30 \times 5 = 150\)

• According to the problem statement, each element in \(S\) belongs to exactly 10 of the \(A_i\) sets. This means if we sum the sizes of all \(A_i\), we are effectively counting each element of \(S\) exactly 10 times.
• So, the total sum (150) must equal 10 times the number of elements in \(S\), denoted as \(|S|\).

  \(150 = 10 \times |S|\)

• By rearranging this equation, we can find the size of \(S\):

  \( |S| = \frac{150}{10} \)

  \( |S| = 15 \)

Therefore, the universal set \(S\) contains 15 elements.

Determining the Value of n

Now we use the information about the \(B_j\) sets and the size of \(S\) we just calculated to find \(n\).

• The sum of the number of elements across all \(n\) sets \(B_j\) is:

  \(\sum_{j=1}^{n} |B_j| = n \times 3 = 3n\)

• The problem states that each element in \(S\) belongs to exactly 9 of the \(B_j\) sets. This means summing the sizes of the \(B_j\) sets counts each element of \(S\) exactly 9 times.
• Thus, the total sum (\(3n\)) must equal 9 times the number of elements in \(S\):

  \(3n = 9 \times |S|\)

• Substituting the value \(|S| = 15\) into the equation:

  \(3n = 9 \times 15\)

  \(3n = 135\)

• Solving for \(n\):

  \(n = \frac{135}{3}\)

  \(n = 45\)

Final Answer

Following the steps using the Principle of Double Counting, we find that the value of \(n\) is 45.