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Primary and secondary routes connecting two computers need to be chosen. Two primary routes are needed from eight which are suitable and three secondary routes must be chosen from four available. In how many ways can the routes be chosen?

Short Answer

Expert verified
There are 112 different ways to choose the primary and secondary routes.

Step by step solution

01

Calculate Primary Route Combinations

First, find the number of ways to choose the two primary routes from the eight available. This is a combination problem, where you're choosing 2 routes from 8, without regard to order. Use the combination formula: \( C(n, k) = \frac{n!}{k! (n - k)!} \), where \( n \) is the total number of options (8 routes) and \( k \) is the number of choices to make (2 routes). So the calculation for primary routes is \( C(8, 2) = \frac{8!}{2! (8 - 2)!} \).
02

Calculate Secondary Route Combinations

Next, find the number of ways to choose the three secondary routes from the four available. Again, this is a combination problem, where you're choosing 3 routes from 4. Use the combination formula: \( C(n, k) = \frac{n!}{k! (n - k)!} \), with \( n = 4 \) and \( k = 3 \). So the calculation for secondary routes is \( C(4, 3) = \frac{4!}{3! (4 - 3)!} \).
03

Find Total Number of Routes

Finally, since the choice of primary and secondary routes are independent events, multiply the number of ways to choose the primary routes by the number of ways to choose the secondary routes to obtain the total number of possible combinations. This is given by \( C(8, 2) \times C(4, 3) \).
04

Perform the Calculations

Calculate the number of combinations for primary and secondary routes using the values from the previous steps: \( C(8, 2) = \frac{8!}{2! (8 - 2)!} = \frac{8 \times 7}{2 \times 1} = 28 \) combinations for primary routes, and \( C(4, 3) = \frac{4!}{3! (4 - 3)!} = \frac{4}{1} = 4 \) combinations for secondary routes. Then multiply these results: \( 28 \times 4 = 112 \) possible combinations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combination Formula
When faced with the task of selecting items from a larger group without caring for the order in which they're chosen, the combination formula comes into play. It's a fundamental concept in combinatorics, a branch of mathematics dedicated to counting, arranging, and grouping items.

The formula is expressed as: \( C(n, k) = \frac{n!}{k! (n - k)!} \), where \( n \) is the total number of items to choose from, \( k \) is the number of items to be chosen, and the symbol \( ! \) denotes a factorial. To provide a practical understanding:
  • Imagine you're at a buffet with 8 dishes, and you want to try only 2. How many ways can you create your plate?
  • Or, picture a soccer team of 11 players, but only 5 can be forwards. The combination formula will help you determine the number of possible forward lineups.
Such problems are efficiently solved using the combination formula because it effortlessly calculates the possible groupings, omitting permutations or different orderings which do not matter in these scenarios.
Factorial Notation
Factorial notation is a compact way to represent the multiplication of a series of descending natural numbers and is essential in calculations involving permutations and combinations. The notation \( n! \) (read as 'n factorial') means multiplying all whole numbers from \( n \) down to 1. For example:
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \( 3! = 3 \times 2 \times 1 = 6 \)
Notably, by convention, \( 0! \) is defined as 1, which often confuses beginners. Factorials grow rapidly with larger numbers, making manual calculations unwieldy, which is why understanding factorial notation is crucial for simplifying such calculations, especially in combinatorial problems.
Independent Events in Probability
In probability, events are considered independent when the outcome of one event does not influence the outcome of another. This concept is pivotal for analyzing scenarios where multiple choices or occurrences are involved.

For example, rolling a die and flipping a coin are independent events; the result of the coin toss does not affect what number the die will show. Mathematically, if the events are independent, the probability of both occurring is the product of their individual probabilities: \( P(A \text{ and } B) = P(A) \times P(B) \).
This principle also extends to combination problems. In our exercise, selecting primary and secondary routes are independent choices—the selection of one does not restrict the other. Hence, we multiply the number of combinations of each to find the total possibilities.
Mathematical Problem-Solving
Solving mathematical problems effectively often requires a structured approach. The step-by-step method showcased in the exercise is a classic example of mathematical problem-solving.
It starts by defining the problem clearly, then breaking down the problem into manageable chunks — in this case, separate calculations for primary and secondary route combinations. Afterward, it proceeds to apply appropriate mathematical formulas and principles, such as the combination formula and the concept of independent events. Lastly, the step-by-step process concludes with carrying out the calculations to arrive at the final answer.
Mathematical problem-solving isn't just about getting the answer, but also understanding and applying concepts to new and varied problems, reinforcing the learning and making it applicable to real-world situations.

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