Question

How many arrangements are possible for the letters of the following words: (a) Tennessee (b) Mississippi (c) Kansas (d) Oregon. (e) Manifoba (f) Vesitine

Short answer

(a) 3780, (b) 34650, (c) 360, (d) 720, (e) 40320, (f) 10080

Step-by-step solution

Identify Total Letters and Frequency

Each word is examined to determine its total number of letters and the frequency of each letter. For example, Tennessee has 9 letters with the following frequencies: T=1, E=4, N=1, S=1.

Use Permutation Formula for Repeated Letters

For permutations of words with repeated letters, use the formula: \[ \frac{n!}{p_1! \times p_2! \times ... \times p_k!} \] where \(n\) is the total number of letters, and \(p_1, p_2, ..., p_k\) are the frequencies of each repeated letter.

Calculate Arrangements for Each Word

Apply the formula to each word: (a) Tennessee: \( \frac{9!}{1! \times 4! \times 1! \times 3!} = 3780 \)(b) Mississippi: \( \frac{11!}{1! \times 4! \times 4! \times 2!} = 34650 \)(c) Kansas: \( \frac{6!}{1! \times 2! \times 1! \times 2!} = 360 \)(d) Oregon: \( \frac{6!}{1! \times 1! \times 1! \times 1! \times 1! \times 1!} = 720 \)(e) Manifoba: \( \frac{8!}{1! \times 1! \times 1! \times 1! \times 1! \times 1! \times 1! \times 1!} = 40320 \)(f) Vesitine: \( \frac{8!}{1! \times 2! \times 2! \times 1! \times 1! \times 1!} = 10080 \).

Key concepts

Permutations

Permutations in combinatorics refer to different ways in which a set of items can be arranged. When we talk about permutations, it is essential to note that the order of items is vital. This means "ABC" is different from "CAB." For instance, if you have three distinct letters, like A, B, and C, you can arrange them in various ways: ABC, ACB, BAC, BCA, CAB, and CBA. This gives exactly 6 permutations, calculated by the formula for permutations of distinct items:

• The number of permutations of a set of items is represented by the factorial of the number of elements, denoted as \( n! \).
• In our example with three letters, this is \( 3! = 3 \times 2 \times 1 = 6 \).

Breaking down permutations provides insight into how many different sequences can emerge from a given set, especially crucial when calculating arrangements of letters in words.

Factorials

Factorials play a crucial role in calculating permutations. A factorial, denoted as \( n! \), represents the product of all positive integers up to \( n \). For example, \( 5! \) means \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials are used to determine how many ways you can arrange a subset of items from a larger set.

• If a word has n letters, the possible arrangements without considering any repetition is \( n! \).
• For words like "Tennessee," calculating its total permutations involves using its total letter count: \( 9! \) for nine letters, which results in 362,880 if no letters repeat.

Factorials grow rapidly and can handle calculations of arrangements in larger sets, enabling us to solve real-world problems where order is a factor.

Repeated Letters

When letters repeat in a word, the calculation of permutations changes to account for this repetition. Simply using \( n! \) would overcount the permutations since rearranging identical letters does not result in a new arrangement. Instead, we use a specific formula for words with repeated letters:

• The formula is: \[ \frac{n!}{p_1! \times p_2! \times ... \times p_k!} \]
• Here, \( n \) is the total number of letters, and \( p_1, p_2, ..., p_k \) are the factorials of the frequencies of each repeated letter.

For example, in the word "Mississippi," the frequencies are significant: four 'S's, four 'I's, and two 'P's, affecting how permutations are counted. By dividing by the repeated letter factorials, we eliminate permutations accounted for with repeating letter rearrangements, reflecting actual unique permutations.

Arrangements

Arrangements refer to different ways in which elements, such as letters, can be sorted or ordered. When dealing with words that may have repeated letters, calculations adjust to reflect unique arrangements by using permutations and factorials.

• With unique letters, the number of arrangements is simply \( n! \).
• However, with repeated letters, the formula \[ \frac{n!}{p_1! \times p_2! \times ... \times p_k!} \] provides a realistic count of unique sequences.

Let's use "Tennessee" as an example: with nine letters, but several repeats (like four 'E's), the arrangement problem accounts for the symmetry among identical letters, reducing the final count from \( 9! \) to a smaller, accurate number. Understanding arrangements this way helps solve complex problems in fields like cryptography, scheduling, and design where order matters but overlaps of similar items occur.