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Chapter 7: Problem 6

How many five-letter words formed with \(a, b\), and \(c\) have at least one letter missing?

Short Answer

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Step by step solution

01

Determine Total Possible Outcomes

First, calculate the total number of five-letter words that can be formed using the letters 'a', 'b', and 'c' if there are no restrictions on repetitions or missing letters. Each position in the five-letter word has three choices (either 'a', 'b', or 'c'), so the total number of combinations is given by: \[ 3^5 = 243 \]
02

Calculate Words with All Letters Present

Next, calculate how many words include all three letters at least once. Use the principle of inclusion-exclusion:1) There are \(3^5 = 243\) total words.2) Subtract words missing 'a': Treat the positions as only having two options, 'b' or 'c', giving \(2^5 = 32\) words.3) Similarly, subtract words missing 'b': Also results in \(2^5 = 32\) words.4) Subtract words missing 'c': Also results in \(2^5 = 32\) words.5) Add back words doubly counted from missing two letters at once such as 'a' and 'b', 'b' and 'c', or 'a' and 'c'. Each results in \(1^5 = 1\) word as there's only one possibility for that combination.Therefore, the number of words with all letters is:\[ |A \cup B \cup C| = 243 - 32 - 32 - 32 + 1 + 1 + 1 = 149 \] where A, B, and C are the sets of words missing 'a', 'b', 'c' respectively.
03

Calculate Words with At Least One Letter Missing

Subtract the number of words that contain all three letters from the total number of five-letter words. This gives:\[ 243 - 149 = 94 \] This is the count of words with at least one letter ('a', 'b', or 'c') missing.

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Key Concepts

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

Combinatorics
Combinatorics is the area of mathematics that deals with counting, arrangement, and combination of objects. It often involves identifying how many ways we can select or arrange items under certain constraints. For the problem of forming five-letter words from the letters 'a', 'b', and 'c', each position in the word offers three possibilities. This leads to the calculation of the full set of potential outcomes using the formula for permutations where repetition is allowed. Here, considering each position independently, we find the total number of combinations by raising the number of choices per position to the power of the number of positions, i.e., \(3^5\). This method highlights how combinatorics simplifies complex counting scenarios into manageable processes, providing a foundation for solving a wide variety of counting problems.
Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a crucial combinatorial technique for accurately counting the number of elements in the union of multiple sets. It ensures that we correctly account for overlapping elements that may otherwise be counted more than once. In the exercise, we used this principle to find how many five-letter words include all three letters 'a', 'b', and 'c'. We began by finding the total number of unrestricted combinations. Next, we excluded those combinations entirely missing one of the letters, which appear in multiple sub-categories of our total set. We calculated each of these separately and adjusted for overlap. By re-adding the words that appear in these overlaps (those missing two letters at once), we avoided undercounting. This step-by-step application confirms the detailed balancing act of combinatorics, showing the power of inclusion-exclusion in managing comprehensive sets, leading to precise results.
Counting Principles
Counting principles are the rules and techniques used to count objects in systematic and structured ways. Fundamental principles include the product rule, the sum rule, and more complex principles like inclusion-exclusion. In our problem scenario, we employed the product rule for calculating the initial set of all possible five-letter words, counting combinations that obey simple placement rules. Then, using inclusion-exclusion, we managed to refine our count to consider only specific combinations with a desired character composition. These principles allow for efficient and accurate resolution of counting problems, showcasing how assumptions and restrictions are seamlessly integrated to model real-world situations mathematically. Building skills in these foundational principles equips students to tackle a wide range of mathematical challenges with confidence.

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Most popular questions from this chapter

A survey asks the respondent to order by importance 10 properties of a car. How many orderings are possible? How many orderings are there if the first and last property are given?

Internet Addresses: IPv4 and IPv6. The Internet requires an address for each machine that is connected to it. The address space of the addressing architecture of Internet Protocol version 4 (IPv4) consists of a 32 -bit field. Since not every combination of bits can be used as an address, plans are underway to change the address space to a 128 -bit field in IPv6. The 32 -bit IPv4 addresses are usually written in a form called dotted decimal. The 32 bit address is broken up into four 8 -bit bytes, and these bytes are then converted to their equivalent decimal form and separated by dots. For example. $$ \begin{array}{ll} 1000000000000011 & 00000010000000011 \end{array} $$ is written as 128.3 .2 .3 , which is obviously more readable. The 128 -bit IPv6 addresses are divided into eight 16 -bit pieces. Each 16 -bit piece is converted to its equivalent hexadecimal value (each sequence of 4 bits is converted to one hexadecimal digit). The eight four-character hexadecimal strings are separated by colons. It is not prac. tical to list 128 bits and show the conversion to the final IPv6 address form. As an example of what you might end up with, however, we show one IPv6 address: \(\mathrm{FFDC} \cdot \mathrm{BA} 98: 7654 \cdot 3210: \mathrm{FEDC}: \mathrm{BA} 98: 7654 \cdot 3210\) How many IPv4 addresses are possible?

Prove \(\sum_{k=0}^{n} C(n, k)^{2}=C(2 n, n)\) (a) Prove the identity using the fact that \((1+x)^{2 n}=(1+x)^{n}(1+x)^{n}\), (b) Give a combinatorial proof of the identity in part a. (c) Find the number of 14 -digit binary sequences for which the number of 1 's in the first seven digits is the same as the number of 0 's in the last seven digits of the sequence. Enumerate all such sequences of length six.

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For \(n=1,2,3, \ldots,\) write $$[x]_{t}=x(x-1)(x-2) \cdots(x-t+1)$$ for \(0 \leq t \leq n .\) We can represent \(\mid x]_{t}\) as a linear combination of powers of \(x .\) The coefficients for this expansion are denoted as \(s(n, t)\) and are known as the Stirling numbers of the first kind. Thus, for any \(n,\) we can write $$[x]_{t}=\sum_{t=0}^{n} s(n, t) x^{t}$$ The numbers \(s(n, t)\) can be defined as \(s(n, 0)=0\) for \(n=1,2,3, \ldots ; s(n, n)=1\) for \(n=0,1,2, \ldots ;\) and $$s(n, t)=s(n-1, t-1)-(n-1) s(n-1, t)$$ for \(t=1,2, \ldots, n-1 .\) Make a table of the Stirling numbers of the first kind for \(n=\) 1,2,3,4,5,6
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