Question

Consider a set of letters consisting of 4 different vowels (including e) and 8 different consonants (including b). How many different 5-letter sequences can be formed using 2 vowels and 3 consonants (with no repeated letters)

(a) containing b?

(b) beginning with b?

(c) beginning with e?

(d) beginning with e and containing b?

Answer 1

Serh Consider Consider a set of letters consisting of a different vowels (including e) and & different consonants (including b) - are 5 different ways that the sequence can contain b There are 5 different for each of these sways there are y vowels and 7 consonants from which 2 vowels and 2 consonant can be chooßen .. The number of ways =" x 7C₂ = 6*21 = 12.6 So total number of ways, 5 letter sequence containing b can be formed using a vowels and 3 consonants - 5x126*4! - 630x24 = 15120 And. (6 Total numbe otal number of ways, 5-letter sequence beginning with b can be formed using 2 vowels and 3 consonants = C₂X²C₂X4! = 6x21x24 - 3024 And Total number of ways, 5-letter sequence beginning with e can be formed wing a rowels and 3 consonants Tel I I X4! =3856824 = 4032 And @ There are 4 different ways that the sequence can begin with e and contain b [elblo Lello D , elbo, el 6] For each of these four cases, there are 3 vowels and 7 consonant from which Ivowel and other a consonant can be choosen. So, the total number of ways, 5-letter bequence containing b and beginning with e can be formed using 2 vowels and 3 consonents=4x (3C, 7(2x31) = (x3x21x6 = 1512 Ans.