Problem 13 - Algebra and Trigonometry

Question

Find the number of distinguishable permutations of the given letters. $$ A A B C D $$

Step-by-Step Solution

Answer

The number of distinguishable permutations is 60.

Step 1: Identify Total Letters

First, count the total number of letters given, which are A, A, B, C, and D. This makes a total of 5 letters.

Step 2: Apply Permutation Formula

To find the number of distinguishable permutations, we use the formula: \[ \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \] where \( n \) is the total number of letters, and \( n_1, n_2, \ldots, n_k \) are the frequencies of the repeated letters.

Step 3: Calculate Factorials

The total number of letters \( n = 5 \). Calculate \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Step 4: Count Repeated Letters

The letter A is repeated twice. Hence, the factorial for the repeated letter A is \( 2! = 2 \times 1 = 2 \).

Step 5: Substitute into Formula

Substitute the values into the permutation formula: \[ \frac{5!}{2!} = \frac{120}{2} = 60 \]. This accounts for the repetition of A.

Key Concepts

Permutation Formula

The permutation formula allows us to calculate the number of different ways we can arrange a set of items. When dealing with letters, this means rearranging them to form new sequences. However, if some of these letters are identical, we need a modified version of the permutation formula. This version considers these repetitions and provides the number of distinguishable, or unique, permutations.

The basic formula for permutations is:

• For arranging n distinct items: \( n! \) (n factorial)

When repetitions occur, we modify it to:

• \( \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \)

Here, \( n \) is the total number of items and \( n_1, n_2, \ldots, n_k \) are the counts of each repeated item.
In our example, since 'A' is repeated, we apply this formula to account for these similarities.

Factorial Calculation

A factorial, denoted by the exclamation mark (!), is a product of an integer and all the integers below it down to one. It's a way to represent the total number of possible arrangements or sequences for a set of distinct items.

For example, \( 5! \) (Read as 'five factorial') is calculated as:

• \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Factorials are fundamental when using permutations, as they help calculate the total arrangements without repetition. In our exercise, the factorial of the total number of letters (5) was computed as 120.
This value represents how many ways you can arrange five different letters if they were all unique.

Repeated Letters

Repeated letters pose a unique challenge in permutation problems because simply counting all arrangements would overestimate the number of unique sequences. When we have letters like 'A' in 'A A B C D' appearing more than once, not all permutations create a different sequence.

To handle this, we calculate a factorial for the count of each repeated letter. In our example:

• The letter 'A' is repeated 2 times, so we compute: \( 2! = 2 \times 1 = 2 \)

By dividing the total permutations (\( 5! \) in the example) by the factorial of the repeated letters (\( 2! \)), we eliminate duplicate sequences.
This ensures that the final count reflects only the distinct permutations, giving us 60 unique arrangements for 'A A B C D' when accounting for the repeated 'A's.