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Craig invited four friends to watch a TV show. He arranged 5 seats in a row. The number of ways he and his four friends can sit in the row is \(n\). In how many of these ways can Craig sit in the middle? (A) \(n\) (B) \(n / 2\) (C) \(n / 3\) (D) \(n / 4\) (E) \(n / 5\)

Short Answer

Expert verified
The answer is (E) \( \frac{n}{5} \).

Step by step solution

01

Determine Total Arrangements

First, find the total number of ways Craig and his four friends can sit in the row. There are 5 people to arrange in 5 seats. The number of possible arrangements is given by 5 factorial, expressed mathematically as \( 5! \). Calculate \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Thus, \( n = 120 \).
02

Fix Craig in the Middle

To find the number of arrangements where Craig sits in the middle, we fix Craig in the middle seat. After placing Craig, there are 4 seats left for his four friends.
03

Arrange Remaining Friends

There are 4 remaining friends, and 4 seats to arrange them in. The number of ways to do this is given by 4 factorial, or \( 4! \). Calculate \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).
04

Compute Proportion

To find the fraction of arrangements where Craig is in the middle, divide the number of ways Craig can sit in the middle (\( 24 \)) by the total number of arrangements (\( 120 \)). This gives \( \frac{24}{120} = \frac{1}{5} \). Hence, the number of ways Craig sits in the middle is \( \frac{n}{5} \).

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Key Concepts

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

Factorial Calculation
Factorials are a core part of combinatorics, especially when calculating permutations and combinations. To put it simply, a factorial of a number, say 5, denoted as \(5!\), is the product of all positive integers from 1 to that number. So, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). This operation helps determine how to arrange things completely in different scenarios. This fundamental operation appears frequently in problems where you need to arrange a set of items, like people in seats.

When faced with a problem requiring arrangement or order, factorial calculation is the tool that tells us how many distinct ways items can be organized. Think of it like figuring out the number of ways to shuffle cards in a deck or arrange books on a shelf with a fixed order.
Permutations
Permutations revolve around arranging a set of items in a specific order. The order matters in permutations, unlike combinations where order does not matter. In the given problem, we need to determine the number of ways to arrange Craig and his four friends in a row of seats, which is a permutation problem.

For instance, when we calculate the total number of ways to arrange 5 people in 5 seats, we use \(5!\). However, if Craig's position is fixed, like when he always sits in the middle, then we focus on arranging the remaining friends. Suppose there are 4 seats remaining for 4 friends, the number of permutations is \(4!\), which is \(24\) arrangements as calculated in the step-by-step solution.
  • Order matters: Different seating orders count as different permutations, which is why calculations involve factorials.
  • Fixed positions: Fixing one item, like Craig in the middle, reduces the permutation to the other items.
Mathematical Problem Solving
Mathematical problem solving involves breaking down a problem into manageable parts and applying logical reasoning and operations to solve it. In combinatorics, it means identifying when to use permutation and factorial calculations to find solutions.

In this particular problem, we first calculated the total number of seating arrangements using factorials because each configuration of 5 people in 5 seats is a permutation. After understanding this, we fixed a condition for Craig sitting in the middle. This strategic step converted a general arrangement problem into a more specific one.
Then, by determining the arrangement of the remaining friends, the solution narrows through targeted operations like the second factorial calculation. Lastly, simplifying the specific condition by evaluating proportions guided us to notice that Craig's fixed middle arrangement forms one-fifth of the total configurations. Such techniques are at the heart of cracking combinatorial problems efficiently.

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