help ~~~

Question

Assuming repetition is allowed. Find the number of strings of five English letters that
(a)contain no vowels
(b)start with a vowel
(c)at least one vowel
(d)begin with B or end with H
(e)start with C and contain at least one vowel

Answer 1

a)C(contain no vowels)=C(the first is consonant)xC(the second is consonant)xC(the third is consonant)xC(the fourth is consonant)xC(the fifth is consonant)

=21x21x21x21x21=21^5=4084101

b)C(start with a vowel)=C(the first is vowel)xC(the second to fifth are consonant)
=5x2^4=972405

c)C(at least one vowel)=All combination-C(conatin no vowels)
=26^5-21^5=7797275

d)C(begin with B or end with H)
=C(the word begins with B)+C(the word end with H) -C(begin with B and end with H)
=(1)x(26^4)+(26^4)x(1)-(26^3)
=896376

e)C(start with C and contain at least one vowel)
=C(first letter is C)xC(the remaining letters contain at least one vowel)
=(1)x(all combination of 4 letters - C(the remaining 4 letters contain no vowels)
=26^4-21^4=262495

Answer 2

(a)
there are 26 letters and 5 vowels (a,e,i,o,u) also repetition is allowed
the number of strings of five English letters that contain no vowels
=21^5
(b)
the number of strings of five English letters that start with a vowel
=5*26^4
(c)
the number of strings of five English letters that at least one vowel
=total number of strings of five English letters- the number of strings of five English letters that contain no vowels
=26^5-21^5
(d)
the number of strings of five English letters that begin with B or end with H
=26^4+26^4
=2(26^4)
(e)
the number of strings of five English letters that start with C and contain at least one vowel
=total number of strings of five English letters that start with C- the number of strings of five English letters that that start with C and contain no vowels
=26^4-21^4

2006-11-27 22:57:53 補充：
(d) is wrongthe number of strings of five English letters that begin with B or end with H =26^4 26^4-26^3=2(26^4)-26^3