Problem 29 - Algebra for College Students

Question

How many different nine-letter permutations can be formed from three identical A's, four identical B's, and two identical C's? 1260

Step-by-Step Solution

Answer

There are 1,260 different nine-letter permutations.

Step 1: Recognize the problem type

This problem involves finding the number of permutations of a multiset, where some elements are repeated. Specifically, there are multiple identical letters (A's, B's, and C's) in the set.

Step 2: Use the permutation formula for multisets

The formula to find the number of permutations of a set with repeated items is:\[\text{Permutations} = \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}\]where \(n\) is the total number of items, and \(n_1, n_2, \ldots, n_k\) are the counts of each distinct item.

Step 3: Identify the quantities to substitute

In this problem, the total number of letters \(n\) is 9 (three A's, four B's, and two C's). The counts for the permutations formula are:- A's: 3- B's: 4- C's: 2.

Step 4: Compute the factorials

Calculate the necessary factorials:- \(9! = 362,880\)- \(3! = 6\)- \(4! = 24\)- \(2! = 2\).

Step 5: Plug into the permutation formula

Substitute into the formula:\[\frac{9!}{3! \times 4! \times 2!} = \frac{362,880}{6 \times 24 \times 2}\]Calculate the denominator:\[6 \times 24 \times 2 = 288\]

Step 6: Perform the division

Complete the calculation:\[\frac{362,880}{288} = 1,260\]

Step 7: State the result

The number of different nine-letter permutations that can be formed is 1,260.

Key Concepts

Factorials

Understanding factorials is essential when working with permutations. A factorial, denoted by the symbol "!", is the product of all positive integers up to a certain number. For example, the factorial of 5, written as 5!, is calculated as:

• 5! = 5 × 4 × 3 × 2 × 1 = 120

Factorials are used in many areas of combinatorics to find the total possible arrangements when order matters. They allow us to deal with large numbers that indicate the vast number of combinations possible in various scenarios. For instance, in the expression 9!, there are 362,880 different ways to arrange 9 distinct items. Factorials grow very rapidly, which is why they help in calculating permutations and combinations efficiently.

Identical Items

In certain permutations, not all items are unique. When dealing with identical items, we must adjust our calculations to avoid overcounting arrangements that are indistinguishable from others. Consider the example with three identical A's, four B's, and two C's. Here:

• The identical A's mean that swapping any two of them results in the same arrangement.
• Similarly, swapping any of the identical B's or C's results in a repeated permutation.

To account for this, we use a permutation formula that divides by the factorial of the number of identical items. This effectively cancels out the extra count created by identical swapping.

Multisets

A multiset is a set that allows for multiple instances of the same element. Unlike traditional sets where each element is unique, multisets can contain repeated items. In our problem, the collection of letters such as three A's, four B's, and two C's form a multiset. Multisets are common in problems involving permutations with identical items. They require special consideration because the number of arrangements depends on both the total number of elements and the specific count of each element type. Handling multisets involves understanding both the total permutations and the adjustments needed to account for repeats.

Permutation Formula

The permutation formula for multisets is a powerful tool to find the number of unique permutations when some items are identical. It is given by:\[\text{Permutations} = \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}\]where \(n\) is the total number of items, and \(n_1, n_2, \ldots, n_k\) are the counts of each distinct item within the set.
Using this formula, we can determine how repeated items affect the total number of possible arrangements. To apply the formula:

• Calculate the factorial of the total number of items \(n!\).
• Calculate the factorial for each distinct group of identical items.
• Divide the total factorial by the product of the factorials of identical item counts to find the unique permutations.

In this problem, this approach simplifies the counting by factoring out equivalent rearrangements, resulting in a concise answer to what might otherwise be a complex counting task.