Problem 40
Question
A recent survey by the MAD corporation indicates that of the 700 families interviewed, 220 own a television set but no stereo, 200 own a stereo but no camera, 170 own a camera but no television set, 80 own a television set and a stereo but no camera, 80 own a stereo and a camera but no television set, 70 own a camera and a television set but no stereo, and 50 do not have any of these. Find the number of families with: At least one of the items.
Step-by-Step Solution
Verified Answer
The number of families with at least one of the items (television set, stereo, or camera) is 820. However, there seems to be an error in the input data, as the total number of families surveyed should be 700.
1Step 1: Understand the given information
We are given the following information about the families and the items they own:
- 220 families own a television set but no stereo (T only)
- 200 families own a stereo but no camera (S only)
- 170 families own a camera but no television set (C only)
- 80 families own a television set and a stereo but no camera (TS only)
- 80 families own a stereo and a camera but no television set (SC only)
- 70 families own a camera and a television set but no stereo (CT only)
- 50 families do not have any of these items (None)
We are asked to find the number of families that own at least one of the items.
2Step 2: Determine the number of families owning one item only, using the Principle of Inclusion-Exclusion
First, let's find the number of families with at least one of the items by adding up the number of families in each category. Using the Principle of Inclusion-Exclusion:
\(Families_{AtLeastOne} = Tonly + Sonly + Conly + TSonly + SConly + CTonly\)
\(= 220 + 200 + 170 + 80 + 80 + 70\)
\(= 820\)
3Step 3: Calculate the total number of families surveyed
To make sure we haven't made any mistakes or double-counted any families, let's calculate the total number of families surveyed. We know that 700 families were interviewed, and 50 of them don't own any of the items:
\(TotalFamilies_{surveyed} = Families_{AtLeastOne} + Families_{None}\)
\(= 820 + 50\)
\(= 870\)
Here, we have 870 families, which is greater than the number of families surveyed (700). This indicates that there must be some mistake in the input data.
However, based on the given data, there are 820 families that own at least one of the items among television set, stereo, or camera.
Key Concepts
Set TheoryCombinatoricsProblem Solving
Set Theory
Set theory is a branch of mathematical logic that studies sets, which are collections of objects. In this context, sets are used to categorize families based on the items they own: television sets, stereos, and cameras.
Each distinct group from the survey can be visualized as subsets within a larger set of all families.
For instance, the subset of families that own only televisions is distinct from those owning only stereos.
When working with sets, we often look at the union, intersection, or difference among them.
- **Union** combines all objects from the involved sets, capturing families owning at least one item. - **Intersection** represents families owning more than one item, like those owning both televisions and stereos.
- **Difference** subtracts one set from another, identifying families owning one item but not another, such as televisions without stereos. Set theory helps us understand how different categories overlap and interact, critical for applying the Principle of Inclusion-Exclusion effectively.
Each distinct group from the survey can be visualized as subsets within a larger set of all families.
For instance, the subset of families that own only televisions is distinct from those owning only stereos.
When working with sets, we often look at the union, intersection, or difference among them.
- **Union** combines all objects from the involved sets, capturing families owning at least one item. - **Intersection** represents families owning more than one item, like those owning both televisions and stereos.
- **Difference** subtracts one set from another, identifying families owning one item but not another, such as televisions without stereos. Set theory helps us understand how different categories overlap and interact, critical for applying the Principle of Inclusion-Exclusion effectively.
Combinatorics
Combinatorics is a field of mathematics focused on counting, arrangement, and combination principles. In this problem, combinatorics comes into play in how we count the number of families owning different combinations of items.
The Principle of Inclusion-Exclusion (PIE) is a key concept in combinatorics used to calculate the size of the union of multiple sets while correcting for over-counting.
With PIE, we consider:
The Principle of Inclusion-Exclusion (PIE) is a key concept in combinatorics used to calculate the size of the union of multiple sets while correcting for over-counting.
With PIE, we consider:
- Adding families who own individual items.
- Subtracting those counted in overlaps more than once (like stereos and televisions).
- Adjusting counts to ensure families with multiple items aren't repeatedly added.
Problem Solving
Problem-solving in mathematics often involves a structured method to reach a solution. For this problem, identifying what is asked – to find families with at least one of the items – is the first step.
Next is organizing known information: families owning televisions without stereos, those with stereos and no cameras, and so forth.
Clarifying such details helps in systematic application of mathematical principles.
Using the given data effectively is key:
Next is organizing known information: families owning televisions without stereos, those with stereos and no cameras, and so forth.
Clarifying such details helps in systematic application of mathematical principles.
Using the given data effectively is key:
- Determine single-item family counts.
- Utilize PIE to prevent over-counting while verifying against total numbers.
- Identify and rectify any discrepancies, such as incorrect totals indicated by more families than surveyed.
Other exercises in this chapter
Problem 39
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