Answer to Question #261055 in Discrete Mathematics for Alina

Question

The English alphabet contains 21 consonants and 5 vowels. How many strings of five lowercase letters can be formed using the following constraints? Give two answers for each of the following - one where repetition is allowed in the string and one where repetition is not allowed.

(a) Only one vowel (placed anywhere)

(b) Maximum two consonants (placed anywhere)

(c) Starts with x, y or z

Answer 1

Solution:

(a)

Five letters with one vowel. We can place the vowel in any of the 5 positions in 5 * 4 ways and the remaining 4 positions can be filled in in 21⁴ ways by the consonants as they can be repeated also. So, total number of words possible if repetition is not allowed $=21^4\times5\times4=3889620$

And total number of words possible if repetition is allowed $=21^4\times5\times5=4862025$

(b)

Maximum two consonants will give (vowels cannot be repeated)

= Zero consonant, five vowels + one consonant, 4 vowels + two consonants, 3 vowels

$=5!+21\times5\times4\times3\times2+21^2\times5\times4\times3 \\=29100$

Maximum two consonants will give (if vowels can be repeated)

= Zero consonant, five vowels + one consonant, 4 vowels + two consonants, 3 vowels

$=5^5+21\times5^4+21^2\times5^3 \\=71375$

(c)

Starts with x, y or z (if repetition is not allowed)

$=3\times25\times24\times23 \\=41400$

Starts with x, y or z (if repetition is allowed)

$=3\times25^3 \\=46875$