Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

How many three-digit numbers can be formed from the digits \(1,2,3,4\), and 5 if repetitions (a) are not allowed? (b) are allowed?

Step-by-Step Solution

Answer

(a) 60 numbers without repetition; (b) 125 numbers with repetition.

Step 1: Identify the available digits

We are given the digits \(1, 2, 3, 4, 5\) to form three-digit numbers. These are the digits we can use in both parts of this exercise.

Step 2: Calculate number of permutations without repetition

For part (a), where repetitions are not allowed, choose 3 different digits from the 5 available. The first digit can be chosen in 5 ways, the second in 4 ways, and the third in 3 ways, leading to the calculation: \(5 \times 4 \times 3 = 60\) three-digit numbers.

Step 3: Calculate number of combinations with repetition

For part (b), where repetitions are allowed, each digit can be chosen independently. Therefore, all three positions (hundreds, tens, and units) can be filled by any of the 5 digits, leading to the calculation: \(5 \times 5 \times 5 = 125\) three-digit numbers.

Key Concepts

Understanding Permutations

Permutations refer to the arrangement of objects in a specific order. When dealing with permutations, the order in which you arrange the items matters, and choosing different orders counts as different permutations.
In our exercise, forming three-digit numbers from the given digits without repetition is a permutation problem. Since each digit must be different and ordered distinctly, we cannot use the digits more than once. Here's how we approached it:

• The first position (hundreds place) can be filled by any of the 5 digits, giving 5 options.
• Once the first digit is used, we have 4 choices left for the tens place.
• For the last digit (units place), we have 3 options left.

Therefore, the number of permutations is calculated as \(5 \times 4 \times 3 = 60\). Each choice impacts the next, making the arrangement unique.

Exploring Combinations

Unlike permutations, combinations focus on selecting items where order does not matter. While this doesn't directly relate to our exercise, it's essential to understand when working through similar problems.
In combinations, we are only concerned with the selection itself, not the arrangement. If we were asked how many sets of three digits could be picked from the digits 1 to 5, regardless of order, it would be a combination problem. For instance, choosing the digits 1, 2, and 3 is the same set as choosing 3, 2, and 1. Since order doesn't matter, such instances don't multiply the number of outcomes.
In general, when calculating combinations where repetition is allowed, the formula differs from when repetition is restricted. However, our exercise requires a focus on arrangements (permutations) rather than selections (combinations).

Considering Repetition in Combinatorics

Repetition plays a significant role in combinatorics. It changes the approach and the outcome of problems significantly. In situations where repetitions are allowed, each choice is independent of the others.
In our exercise's second part, repetition means that each digit can be chosen and used again in forming the number. This changes the method of calculating possible numbers:

• Each position of hundreds, tens, and units can each be filled by any of the 5 digits.
• Thus, we calculate this by multiplying the options for each position: \(5 \times 5 \times 5 = 125\).
  Here, repetition allows total freedom in choice, meaning digit reuse leads to a greater number of possible outcomes.
  Recognizing when repetition is involved helps in setting up calculations and understanding problem requirements effectively.