Problem 34 - Algebra for College Students

Question

Make a list of all of the permutations of the letters \(A, B, C\) \(\mathrm{D},\) and \(\mathrm{E}\) taken 3 at a time. How many permutations should be in your list?

Step-by-Step Solution

Answer

There are 60 permutations of the letters A, B, C, D, and E taken 3 at a time.

Step 1 - Understand Permutations

Permutations are arrangements of items in a specific order. We need to find all permutations of the letters A, B, C, D, and E taken 3 at a time.

Step 2 - Calculate the Number of Permutations

The formula for permutations of n items taken r at a time is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \] Here, n = 5 and r = 3. So, we need to calculate P(5, 3).

Step 3 - Evaluate the Factorials

Calculate the factorials needed: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] \[ (5-3)! = 2! = 2 \times 1 = 2 \]

Step 4 - Compute P(5, 3)

Substitute the values into the formula: \[ P(5, 3) = \frac{5!}{(5-3)!} = \frac{120}{2} = 60 \] Thus, the number of permutations is 60.

Step 5 - List All Permutations

List all the permutations by picking 3 distinct letters from A, B, C, D, and E. For each combination of 3 letters, list all the possible orders: For example: ABC, ABD, ABE, ACB, ACD, ACE, etc. The total number of such arrangements is 60, as computed.

Key Concepts

Permutations Formula

Permutations refer to different ways of arranging a certain number of items in a specific order. The permutations formula helps us determine how many of these arrangements are possible.
When we talk about permutations, order matters. For example, ABC is different from ACB.
If you want to find the number of permutations of n items taken r at a time, you can use the permutations formula:
\[ P(n, r) = \frac{n!}{(n-r)!} \]
Here, \(P(n, r)\) is the number of permutations, \(n!\) is the factorial of the total number of items, and \((n-r)!\) is the factorial of the difference between the total items and the items taken.
Let's consider our exercise where we have 5 letters: A, B, C, D, and E. We want to find the permutations taken 3 at a time.
Plugging in the numbers, we get:
\( P(5, 3) = \frac{5!}{(5-3)!} = \frac{120}{2} = 60 \). So, there are 60 different ways to arrange 3 letters out of the 5.

Factorials

Factorials are a key concept in permutations and combinations. A factorial, denoted by \(!\) (e.g., \(5!\)), is the product of all positive integers up to that number.
Factorials are used because they represent the number of ways to arrange a set of items.
Here's how you calculate factorials step-by-step:

• The factorial of 1 is 1 (\(1! = 1\)).
• The factorial of 2 is \(2! = 2 × 1 = 2\).
• The factorial of 3 is \(3! = 3 × 2 × 1 = 6\).
• And so on.

In our exercise, we need to calculate \(5!\) and \( (5-3)!\).
For \( 5! \):
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
For \( (5-3)! \):
\[ 2! = 2 \times 1 = 2\]
These factorials are then used in the permutations formula:
\[ P(5, 3) = \frac{5!}{(5-3)!} = \frac{120}{2} \].

Combinatorial Arrangements

Combinatorial arrangements are different ways we can select and arrange a subset of items from a larger set.
In our exercise, we're looking at permutations, which means we care about the order of selection.
Let's break down how we list all possible permutations of 3 letters chosen from 5: A, B, C, D, and E:
You can start by picking any of the 5 letters as the first letter. Then, pick any of the remaining 4 letters as the second letter. Finally, pick any of the remaining 3 letters as the third letter.
For example, if we pick A first, we have 4 choices for the second letter (B, C, D, E), and 3 choices left for the third letter:

• ABC, ABD, ABE, ACB, ACD, ACE, etc.

Doing this for all possible selections will give us the total permutations. Remember, we calculated 60 permutations using the formula, so there are 60 unique ways to arrange these letters.
Using these concepts can help you solve similar permutation problems efficiently.