Problem 26 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Classify each problem according to whether it involves a permutation or a combination. How many three-digit numbers can be formed using the numerals in the set \(\\{3,2,7,9\\}\) if repetition is not allowed?

Step-by-Step Solution

Answer

This problem involves permutations since the order of the numerals matters. There are 24 three-digit numbers that can be formed using the numerals in the set {3, 2, 7, 9} with no repetition allowed.

Step 1: Identify the problem type

We need to form a three-digit number with non-repeating numerals, so this is a problem involving permutations because the order of the numerals matters.

Step 2: Determine the number of numerals

We have four numerals in the set: 3, 2, 7, and 9.

Step 3: Calculate the permutations

Since repetition is not allowed and we want to create a three-digit number, we need to make permutations of size 3 from the set of 4 numerals. This is calculated as follows: Permutations = \( _nP_r = \frac{n!}{(n-r)!} \) Here, n = 4 (number of numerals in the set) and r = 3 (size of the permutation) Permutations = \( _4P_3 = \frac{4!}{(4-3)!} = \frac{4!}{1!} = \frac{4*3*2*1}{1} \) = 24 There are 24 permutations possible.

Step 4: Conclusion

So, there are 24 three-digit numbers can be formed using the numerals in the set {3, 2, 7, 9} with no repetition allowed. This problem involves permutations as the order of numerals matters.

Key Concepts

Understanding Combinations

Combinations are mathematical selections where the order does not matter. Unlike permutations, you simply choose items without caring about which comes first.

Consider a scenario where you choose 2 fruits out of an apple, banana, and orange. Whether you pick an apple first or an orange, it doesn’t change the group. You use combinations here. The formula is:

• Combinations: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
• \( n \) is the total number of items to choose from.
• \( r \) is the number of items to choose.

Knowing when to use combinations helps in tackling many real-world problems with ease.

Factorials Simplified

The factorial of a number, denoted by \( n! \), is the product of all positive integers up to \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). This concept is pivotal for both permutations and combinations.

Here are some quick tips:

• \( 0! = 1 \), by definition.
• Factorials grow quickly with larger numbers, e.g., \( 5! = 120 \), \( 6! = 720 \).
• Used extensively in probability and statistics.

Understanding factorials helps grasp more complex formulas, like those in permutations and combinations.

Using the Permutation Formula

Permutations focus on arrangements where the order is important. For instance, forming different passwords or seating arrangements requires permutations.

The permutation formula is:

• Permutations: \( _nP_r = \frac{n!}{(n-r)!} \)
• \( n \) is the number of available items.
• \( r \) is the number of items to arrange.

In our exercise, we calculated how many three-digit numbers can be formed from four numbers. Here's the breakdown:

• We had 4 numerals: 3, 2, 7, 9.
• We wanted to create a three-digit number (\( r = 3 \)).
• By calculating \( _4P_3 \), we found 24 possible arrangements.

This shows the importance of sequence in permutations.

Mathematical Problem Solving Strategies

Solving math problems often involves identifying the type of problem and choosing the right formula or method. Here’s a simple approach:

• Identify whether the order matters (permutations) or not (combinations).
• Determine the number of items to arrange or pick.
• Choose the appropriate formula: permutation or combination.

Here’s a quick reminder:

• If you need order, use permutations.
• If you only need selection, use combinations.

By systematically understanding the problem and applying formulas, you make solving such problems a breeze.