Problem 12 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

There are 1680 distinguishable permutations.

Step 1: Understand the Problem

The task requires finding the number of different arrangements (permutations) of the letters A, A, A, B, B, B, C, C, C.

Step 2: Identify Total Number of Letters

Count the total number of letters. We have 9 letters in total: 3 A's, 3 B's, and 3 C's.

Step 3: Apply the Permutation Formula

To find the number of distinguishable permutations of letters where there are repeating elements, use the formula: \[ \frac{n!}{n_1! \times n_2! \times n_3!} \] where \(n\) is the total number of letters, and \(n_1, n_2, n_3, ...\) are the frequencies of each distinct letter. Here, \(n = 9\), \(n_1 = 3\) (A's), \(n_2 = 3\) (B's), \(n_3 = 3\) (C's).

Step 4: Calculate Factorials

Calculate the factorials: \(9! = 362880\) and \(3! = 6\).

Step 5: Substitute into the Formula

Substitute the factorials into the permutation formula: \[ \frac{9!}{3! \times 3! \times 3!} = \frac{362880}{6 \times 6 \times 6} \]

Step 6: Perform the Division

Compute the division: \[ \frac{362880}{216} = 1680 \].

Key Concepts

Permutation Formula

Permutation formulas are mathematical tools used to find the different ways we can arrange a set of objects. When objects in the set are distinct, the formula is straightforward: simply find the factorial of the number of objects.
However, when some objects are identical, as in the case of letters such as A, B, and C in our exercise, we need to adjust the formula to reflect these repeating elements. This is where the distinguishable permutation formula comes in, given by:

• \[ \frac{n!}{n_1! \times n_2! \times \ldots} \]

Here, \(n\) represents the total number of objects, and \(n_1, n_2, \) etc., represent the counts of each identical object in the set.
By dividing the total arrangements \(n!\) by the factorials \(n_1!\), \(n_2!\), etc., we effectively remove duplicate arrangements caused by the repeating elements, thus finding the truly distinct permutations.

Factorials

Factorials are a core concept in permutations and combinations. A factorial of a positive integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(3!\) means multiplying 3 by every positive integer less than it:

• \(3! = 3 \times 2 \times 1 = 6\)
• \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\)

These calculations become significant in the permutation formula, especially when dealing with larger numbers, like computing \(9!\). Factorials grow dramatically with larger \(n\), so having a solid understanding of them is crucial for solving permutation problems efficiently.

Repeating Elements

In permutations, repeating elements can complicate calculations, as seen with the set of letters A, A, A, B, B, B, C, C, C. When elements repeat, some arrangements that look different are actually the same, because swapping identical letters doesn't create a new combination.
To resolve this, the permutation formula includes dividing by the factorial of the number of each repeated element. This division effectively cancels out the surplus theoretical permutations caused by swapping indistinguishable elements.
Thus, for 3 A's, 3 B's, and 3 C's, we divide by \(3!\) for each:

• \(n_1! = 3!\)
• \(n_2! = 3!\)
• \(n_3! = 3!\)
• Calculation: \(6 \times 6 \times 6 = 216\)

This adjustment ensures only unique arrangements are counted.

Arrangements of Letters

Understanding arrangements of letters in permutations involves recognizing both the total number of letters and the impact of their order. In cases without repetition, every order counts as unique. But in scenarios like the exercise, where letters repeat, only distinct sequences matter.
The example, with A, A, A, B, B, B, C, C, C, shows this complexity. Even if letters are visually different due to their position, their identical nature means the arrangements might not be practically different.
Therefore, calculating distinguishable permutations with this understanding ensures we count genuinely unique configurations, emphasizing rational sorting over merely visual difference. This balance between the quantity and type of letters challenges our initial assumptions about combinations, deepening our insight into arranging letters.