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Q31E

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Discrete Mathematics and its Applications
Found in: Page 414
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

The English alphabet contains \(21\) consonants and five vowels. How many strings of six lowercase letters of the English alphabet contain

a) exactly one vowel?

b) exactly two vowels?

c) at least one vowel?

d) at least two vowels?

  1. The required strings are\(122,523,030\).
  2. The required strings are\(72,930,375\).
  3. The required strings are\(223,149,655\).
  4. The required strings are\(100,626,625\).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definitions of Product rule, Permutation, and Combination

Product rule: If one event can occur in \(m\) ways and a second event can occur in \(n\) ways, then the number of ways that the two events can occur in sequence is then \(m \cdot n\).

Definition permutation (order is important):

\(P(n,r) = \frac{{n!}}{{(n - r)!}}\)

Definition combination (order is not important):

\(C(n,r) = \left( {\begin{array}{*{20}{l}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)with \(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\)

Step 2: Calculate the number of strings with exactly one vowel

(a)

There are\(26\)letters in total of which \(21\) are consonants and\(5\)are vowels.

In strings contains\(6\)letters:

Position vowel:\(6\)ways

Vowel:\(5\)ways

Second letter:\(21\)ways(needs to be a consonant)

Third letter:\(21\)ways(needs to be a consonant)

Fourth letter:\(21\)ways(needs to be a consonant)

Fifth letter:\(21\)ways(needs to be a consonant)

Sixth letter:\(21\)ways(needs to be a consonant)

Use the product rule:

\(\begin{array}{c}6 \cdot 5 \cdot 21 \cdot 21 \cdot 21 \cdot 21 \cdot 21 = 6 \cdot 5 \cdot {21^5}\\ = 122,523,030\end{array}\)

Thus, there are\(122,523,030\)strings containing exactly one vowel.

Step 3: Calculate the number of strings with exactly two vowels

(b)

There are\(26\)letters in total of which\(21\)are consonants and\(5\)are vowels.

In strings contains\(6\)letters:

Position vowels: \(\left( {\begin{array}{*{20}{l}}6\\2\end{array}} \right) = \frac{{6!}}{{2!(6 - 2)!}} = 15\) ways(as the order of the two positions of the vowels is not important)

First vowel:\(5\)ways

Second vowel:\(5\)ways

Third letter:\(21\)ways(needs to be a consonant)

Fourth letter:\(21\)ways(needs to be a consonant)

Fifth letter:\(21\)ways(needs to be a consonant)

Sixth letter:\(21\)ways(needs to be a consonant)

Use the product rule:

\(\begin{array}{c}15 \cdot 5 \cdot 5 \cdot 21 \cdot 21 \cdot 21 \cdot 21 = 15 \cdot {5^2} \cdot {21^4}\\ = 72,930,375\end{array}\)

Thus, there are\(72,930,375\)strings containing exactly two vowels.

Step 4: Find the number of strings with at least one vowel

(c)

There are\(26\)letters in total of which\(21\)are consonants and\(5\)are vowels. We are interested in strings contains\(6\)letters.

There are\(26\)possible letters for each letter in the string.

By the product rule:

\(\begin{array}{c}Number\,of\,strings = 26 \cdot 26 \cdot 26 \cdot 26 \cdot 26 \cdot 26\\ = {26^6}\\ = 308,915,776\end{array}\)

When the string contains no vowels, then there are\(21\)possible letters for each letter in the string.

By the product rule:

Strings that do not have no vowels will have at least one vowel.

Number of strings with at least one vowel\( = \)Number of strings\( - \)Number of strings with no vowels

Number of strings with at least one vowel\( = {26^6} - {21^6}\)

Number of strings with at least one vowel\( = 308,915,776 - 85,766,121\)

Number of strings with at least one vowel\( = 223,149,655\)

\(\)

Step 5: Find the number of strings with at least two vowels

(d)

Strings that do not have no vowels nor one vowel will have at least two vowels.

Number of strings with at least two vowels\( = \) Number of strings \( - \) Number of strings with no vowels \( - \) Number of strings with one vowel

Number of strings with at least two vowels\( = {26^6} - {21^6} - 6 \cdot 5 \cdot {21^5}\)

Number of strings with at least two vowels\( = 308,915,776 - 85,766,121 - 122,523,030\)

Number of strings with at least two vowels\( = 100,626,625\)

Most popular questions for Math Textbooks

In this exercise we will count the number of paths in the\(xy\)plane between the origin \((0,0)\) and point\((m,n)\), where\(m\)and\(n\)are nonnegative integers, such that each path is made up of a series of steps, where each step is a move one unit to the right or a move one unit upward. (No moves to the left or downward are allowed.) Two such paths from\((0,0)\)to\((5,3)\)are illustrated here.

a) Show that each path of the type described can be represented by a bit string consisting of\(m\,\,0\)s and\(n\,\,1\)s, where a\(0\)represents a move one unit to the right and a\(1\)represents a move one unit upward.

b) Conclude from part (a) that there are \(\left( {\begin{array}{*{20}{c}}{m + n}\\n\end{array}} \right)\) paths of the desired type.

Show that (p→r)∨(q→r)and are logically equivalent

a) State the binomial theorem.

b) Explain how to prove the binomial theorem using a combinatorial argument.

c) Find the coefficient of x100y101in the expansion of (2x+5y)201 .

How many bit strings of length 10 contain

a) exactly four 1s?

b) at most four 1s?

c) at least four 1s?

d) an equal number of 0s and 1s ?

In how many different ways can five elements be selected in order from a set with five elements when repetition is allowed?
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