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Chapter 5: Problem 55

How many different 10 -letter words (real or imaginary) can be formed from the letters in the word STATISTICS?

Short Answer

Expert verified
50,400

Step by step solution

01

Count the Frequency of Each Letter

Identify the frequency of each letter in the word 'STATISTICS'. S appears 3 times, T appears 3 times, A appears once, I appears twice, and C appears once.
02

Calculate the Total Number of Letters

Confirm that the total number of letters in 'STATISTICS' is 10, which matches the length of the word that needs to be formed.
03

Apply the Multinomial Coefficient Formula

Use the multinomial coefficient formula to calculate the number of different arrangements: \[ \frac{10!}{3!3!1!2!1!} \].
04

Simplify the Factorials

Simplify the expression by calculating the factorials: \[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \] \[ 3! = 6 \], \[ 2! = 2 \], \[ 1! = 1 \].
05

Divide to Find the Number of Arrangements

Calculate the number of different 10-letter words: \[ \frac{3,628,800}{6 \times 6 \times 1 \times 2 \times 1} = \frac{3,628,800}{72} = 50400 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multinomial Coefficient
The multinomial coefficient is like an extension of the binomial coefficient used in combinatorics. It helps us count the number of ways to arrange a set of items where some items are repeated. For example, given the word 'STATISTICS', we have multiple repetitions: 3 S's, 3 T's, 2 I's, 1 A, and 1 C. The formula for the multinomial coefficient is: \[ \frac{n!}{k_1! k_2! \times \times \times k_m!} \]. Here, n is the total number of items (10 in the word 'STATISTICS'), and each k represents the frequency of each repeated item (3 S's, 3 T's, and so on). This formula tells us how to distribute these items into distinct sets without caring for the order.
Factorial Calculation
A factorial, denoted as n!, means you multiply a series of descending natural numbers. For example, \[10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800\]. Factorials grow very quickly, so calculating them for larger numbers can be tricky without a calculator. Another example, \[3! = 3 \times 2 \times 1 = 6\]. Factorials play a vital role in combinatorics and probability theories because they help arrange a number of objects or items.
Letter Frequency Analysis
Analyzing the frequency of letters in a word is the first step in many combinatorial problems. In the word 'STATISTICS', we see different letters repeated various times. This analysis is crucial for setting up our problem:
  • S appears 3 times.
  • T appears 3 times.
  • I appears 2 times.
  • A appears once.
  • C appears once.
By identifying this frequency, we can apply the multinomial coefficient correctly by plugging these values into our formula.
Permutations and Combinations
Permutations and combinations are fundamental concepts in combinatorial counting. Permutations refer to the arrangement of objects in a specific order, while combinations deal with the selection of objects without worrying about the order. When calculating the number of distinct words from 'STATISTICS', we use permutations since we are interested in the different possible orders of the letters. By using the multinomial coefficient, we ensure that permutations with repeated letters are counted correctly. In simpler terms, for permutations where repetition is allowed, the formula adjusts this count to avoid overestimating the possible arrangements.

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Most popular questions from this chapter

Suppose a local area network requires eight letters for user names. Lower- and uppercase letters are considered the same. How many user names are possible for the local area network?

Companies whose stocks are listed on the NASDAQ stock exchange have their company name represented by either four or five letters (repetition of letters is allowed). What is the maximum number of companies that can be listed on the NASDAQ?

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According to the Centers for Disease Control, the probability that a randomly selected citizen of the United States has hearing problems is 0.151. The probability that a randomly selected citizen of the United States has vision problems is \(0.093 .\) Can we compute the probability of randomly selecting a citizen of the United States who has hearing problems or vision problems by adding these probabilities? Why or why not?
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