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Chapter 1: Problem 9

Suppose sets \(\mathrm{A}_{\mathrm{i}}(\mathrm{i}=1,2, \ldots 60)\) each set having 12 elements and set \(\mathrm{B}_{\mathrm{j}}(\mathrm{j}=1,2,3 \ldots . \mathrm{n})\) each set having 4 elements let \({ }^{60} \mathrm{U}_{\mathrm{i}=1} \mathrm{~A}_{1}={ }^{\mathrm{n}} \mathrm{U}_{\mathrm{j}=1} \mathrm{~B}_{\mathrm{j}}=\mathrm{C}\) and each element of \(\mathrm{C}\) belongs to exactly 20 of \(\mathrm{A}_{i}\) 's exactly 18 of \(\mathrm{B}_{j}\) 's then \(\mathrm{n}\) is equal to (a) 162 (b) 36 (c) 60 (d) 120

Short Answer

Expert verified
Our solution seems to have encountered a logical error, and the correct answer is not among the given options.

Step by step solution

01

Setup for Principle of Inclusion-Exclusion

From the problem, we know that the universal set C can be written as \(\displaystyle\bigcup_{i=1}^{60}\mathrm{A_i}=\displaystyle\bigcup_{j=1}^n\mathrm{B_j}\). Now we will set up the PIE formula to determine the total number of elements in C.
02

Apply PIE for sets A and B

By the Principle of Inclusion-Exclusion, we have: \(|\displaystyle\bigcup_{i=1}^{60}\mathrm{A_i}|=|\displaystyle\bigcup_{j=1}^n\mathrm{B_j}|\) Since each set A has 12 elements, and each element of C belongs to exactly 20 of the A sets, we have a total of \(20 \cdot 12\) elements in all sets A combined without considering repetition. Similarly, since each B set has 4 elements, and each element of C belongs to exactly 18 of the B sets, we have a total of \(18 \cdot 4\) elements in all sets B combined without considering repetition.
03

Equate the total number of elements in sets A and B

As we consider the total number of elements in A or B sets without repetition, we can write the equation: \(20 \cdot 12 = 18 \cdot 4 \cdot n\)
04

Solve for n

Now we'll solve the equation for n: \(240 = 72n\) Divide both sides by 72: \(n = \frac{240}{72} = \boxed{3.33}\) Since n cannot be a non-integer number, we must round it up to the nearest integer. Thus, n is equal to 4. Since 4 is not an option in the given choices, we should take a closer look at our PIE application.
05

Reworking the PIE method

The mistake made previously was in counting the elements from sets A and B without considering the intersections (double-counting). Let's fix that: Every one of the 12 elements in each of the sets A is in 5 of the sets B (as there are 20 sets A for each element and 18 sets B for each element, and 60 A's vs n B's). And similarly, every one of the 4 elements in each of the sets B is in 15 of sets A. So, we have: - Total elements from all A's: \(60 \cdot 12\) - Total elements from all B's: \(4n \cdot 4\) - Intersecting elements (double-counted): \(5 \cdot 12 \cdot n\) Now we can set up our PIE equation: \(60 \cdot 12 = 4n \cdot 4 - 5 \cdot 12 \cdot n\)
06

Solve for n again

Simplify and solve for n: \(720 = 16n - 60n\) \(720 = -44n\) Now divide by -44: \(n = \frac{720}{-44}\) \(n = \boxed{-16.36}\) Since n is negative, this can't be the correct solution. This might be due to round-off error in our integer approximations. To avoid such errors, we'll revert to our first correct equation: \(20 \cdot 12 = 18 \cdot 4 \cdot n\). In this case, we made an error when solving for n. Let's recalculate: \(240 = 72n\) \(n = \frac{240}{72} = \boxed{3.33}\) Since n cannot be a non-integer number, there must be a logical error in our steps. The correct answer, therefore, is not among the given options.

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