Problem 72
Question
A shipment of 30 flat screen televisions contains three defective units. In how many ways can a vending company purchase four of these units and receive (a) all good units, (b) two good units, and (c) at least two good units?
Step-by-Step Solution
Verified Answer
The vending company can purchase four units in (a) 17550 ways where all are good, (b) 1053 ways where two units are good and two are defective, and (c) 20528 ways where there are at least two good units.
1Step 1: Calculating combinations for four good units
The number of ways of selecting four good units out of available 27 units is given by the combination formula: \( C(n, r) = \frac{n!}{r!(n-r)!} \), where n is the number of items to choose from, and r is the number of items to choose. This results in \( C(27, 4) = \frac{27!}{4!(27-4)!} = 17550 \).
2Step 2: Calculating combinations for two good units and two defective units
Following the same formula, the number of ways of getting two good units and two defective units out of the total is calculated as follows: \( C(27, 2) * C(3, 2) = (\frac{27!}{2!(27-2)!}) * (\frac{3!}{2!(3-2)!}) = 351 * 3 = 1053 \).
3Step 3: Calculating combinations for at least two good units
Here 'at least two' can mean two, three or four. So, the calculation will be the sum of combinations where two, three or four units are good. Therefore, we're calculating: \( C(27,2)*C(3,2) + C(27,3)*C(3,1) + C(27,4)*C(3,0) = 1053 + 2925 + 17550 = 20528 \).
Key Concepts
CombinatoricsPermutations and CombinationsFactorials in Algebra
Combinatorics
Combinatorics is a field of mathematics concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is closely related to many other areas of mathematics and has many applications ranging from logic to statistical physics, from evolutionary biology to computer science, etc.
One impressive application of combinatorics is in the context of probability where we want to know the number of ways in which a certain event can occur. In the exercise provided, we're looking into the number of ways a vending company can purchase units with certain conditions. This is a classic example of a combinatorial problem where we need to count combinations without regard to the order of selection.
One impressive application of combinatorics is in the context of probability where we want to know the number of ways in which a certain event can occur. In the exercise provided, we're looking into the number of ways a vending company can purchase units with certain conditions. This is a classic example of a combinatorial problem where we need to count combinations without regard to the order of selection.
Permutations and Combinations
When it comes to counting the number of ways items can be arranged or selected, two concepts are key: permutations and combinations. Permutations of a set of elements are all the possible arrangements of those elements, where the order is important, whereas combinations are ways of selecting items from a collection, such that the order of selection does not matter.
In the exercise, we focus on combinations because the order in which the vending company purchases the television units is not important; only the selection matters. Hence, we use the formula for combinations, expressed as \( C(n, r) = \frac{n!}{r!(n-r)!} \), to solve the problem. By plugging in the respective values for selecting good and defective units, we're able to find the number of possible selections that meet the criteria set out in the different parts of the problem.
In the exercise, we focus on combinations because the order in which the vending company purchases the television units is not important; only the selection matters. Hence, we use the formula for combinations, expressed as \( C(n, r) = \frac{n!}{r!(n-r)!} \), to solve the problem. By plugging in the respective values for selecting good and defective units, we're able to find the number of possible selections that meet the criteria set out in the different parts of the problem.
Factorials in Algebra
An essential tool used in permutations and combinations is the concept of a factorial, denoted by an exclamation point (\(!\)). A factorial represents the product of all positive integers up to a given number. For example, \(5!\) is calculated as \(5 \times 4 \times 3 \times 2 \times 1\), which equals 120.
A key aspect of factorial usage in algebra is its role in calculating permutations and combinations. For instance, the denominators and numerators in the combinations formula involve factorials, which vastly simplifies the process of determining the quantity of different arrangements or selections. In the exercise, the combinations to find the number of ways to select the televisions heavily rely on the factorial operation to handle the large numbers involved efficiently.
A key aspect of factorial usage in algebra is its role in calculating permutations and combinations. For instance, the denominators and numerators in the combinations formula involve factorials, which vastly simplifies the process of determining the quantity of different arrangements or selections. In the exercise, the combinations to find the number of ways to select the televisions heavily rely on the factorial operation to handle the large numbers involved efficiently.
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