Problem 61 - Algebra and Trigonometry

Question

Twenty students are arranged randomly in a row for a class picture. Paul wants to stand next to Phyllis. Find the probability that he gets his wish.

Step-by-Step Solution

Answer

The probability is \( \frac{1}{10} \).

Step 1: Calculate Total Arrangements

First, calculate the total number of ways to arrange twenty students in a row. Since order matters, we use the formula for permutations of 20 objects, which is given by \[ 20! \] (20 factorial).

Step 2: Treat Paul and Phyllis as a Single Unit

Consider Paul and Phyllis as a single unit or block. This way, we just need to arrange 19 units instead of 20, where this 'block' contains both Paul and Phyllis. The number of ways to arrange these 19 units is \[ 19! \].

Step 3: Arrange Paul and Phyllis Within Their Block

Within their block, Paul and Phyllis can be arranged in 2 ways (Paul can be standing first or Phyllis can be standing first). This means, for each configuration of the other 18 students, Paul and Phyllis can be swapped 2 ways, which is given by \[ 2! \].

Step 4: Calculate Favorable Arrangements

Multiply the arrangements of the 19 blocks by the arrangements within the blocks to find the total number of favorable arrangements: \[ 19! \times 2! \].

Step 5: Calculate Probability

The probability is given by the ratio of favorable arrangements (Paul and Phyllis standing together) to the total arrangements: \[ \frac{19! \times 2}{20!} = \frac{2}{20} = \frac{1}{10} \].

Key Concepts

Permutations

Permutations are an essential concept in probability and combinatorics. They refer to the number of ways to arrange a set of objects where the order matters. For instance, if you have a group of three students, A, B, and C, there are several ways you could line them up: ABC, ACB, BAC, BCA, CAB, and CBA. Each different sequence is a permutation.

The general formula for calculating permutations of n distinct objects is given by the factorial function, represented as \( n! \). Here, \( n! \) stands for the product of all positive integers up to n. So, for example, \( 3! = 3 \times 2 \times 1 = 6 \).

• In our exercise, we have 20 students, and we need to arrange them in a row. Therefore, the total number of permutations would be \( 20! \), a tremendously large number representing every possible sequence of student arrangements.
• This approach is necessary whenever we care about who stands where, highlighting why "order matters" in permutations.

Factorial

The factorial concept, represented by the symbol \( ! \), is a fundamental mathematical operation used extensively in permutations, combinations, and probability. It involves computing the product of all positive integers up to a specified number n. Understanding factorials will help you to grasp both the simplicity and the vastness of permutations.

• For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Essentially, factorials demonstrate how rapidly the number of permutations grows as more items are added.
• In our exercise, we start with 20 students, and therefore, \( 20! \) accounts for all possible arrangements of these students in a line. It quickly becomes clear how even a change from 19 to 20 increases the number of arrangements exponentially.

Let's not forget about the factorial of smaller numbers, like 2!, which plays a role whenever we deal with tiny permutations, such as deciding the order within a pair. In the exercise, 2! represents the internal swap options between Paul and Phyllis, leading to 2 possible arrangements.

Combinatorial Problems

Combinatorial problems involve finding the number of ways to arrange or select objects, using permutations and combinations. These problems often reveal the underlying structure of real-world scenarios, like seating arrangements or team formations.

In our exercise, we encounter a common combinatorial setup: ensuring two specific individuals are next to each other. To simplify, we treat Paul and Phyllis as a single unit, or block. By doing this, the problem shifts slightly from arranging 20 students to arranging 19 objects, one of which is this special block.

• This strategy is quite effective whenever groups need to be kept together in permutations. Here, Paul and Phyllis swap back and forth in their block, adding permutations to the block's arrangement.
• The outcome is calculated by multiplying the ways to arrange the blocks by the ways to arrange Paul and Phyllis within their block, producing the favorable outcomes where they are together.

To find the probability, we then compare these favorable outcomes with the total number of permutations, helping us understand the likelihood of Paul getting his wish. By doing this, combinatorial problems like ours become much more manageable and logically structured.