The English alphabet has 26 letters of which 5 are vowels. Consider only 5-
letter “words” consisting of 3 different consonants and 2 different vowels. Find
the number of such words which:
(a) have no restrictions;
Choose the three consonants 21C3=1330 ways
Choose the two vowels 5C2=10 ways
Permute each in 5P5=5!=120 ways
That's 1330∙120=1596000 ways
(c) contain the letters B and C;
Choose the third consonants 19C1=19 ways
Choose the two vowels 5C2=10 ways
Permute each in 5P5=5!=120 ways
That's 19∙10∙120=22800 ways
(b) contain the letter B;
Choose the remaining 2 consonants 20C2=190 ways
Choose the two vowels 5C2=10 ways
Permute each in 5P5=5!=120 ways
That's 190∙10∙120=228000 ways
(d) begin with B and contain the letter C.
Choose the other consonant in 19C1=19 ways.
Choose the two vowels in 5C2=10 ways.
Permute the 4 letters on the right of the B in 4P4=4!=24 ways.
That's 19∙10∙24=4560 ways.
Edwin
