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Q31E

Expert-verifiedFound in: Page 414

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**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?**

- The required strings are\(122,523,030\).
- The required strings are\(72,930,375\).
- The required strings are\(223,149,655\).
- The required strings are\(100,626,625\).

**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 **\(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\)

(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.

(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.

(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\( = 223,149,655\)

\(\)

(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\( = 100,626,625\)

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