In this problem a word is a nite sequence of letters from a

In this problem, a word is a (nite) sequence of letters from a given alphabet, not necessarily a word found in a dictionary. Consider the alphabet A,B,C,D,E,F.
a. How many four-letter words contain the subword ACE?

b. How many four-letter words don’t begin with F or don’t end in E?

c. How many ve-letter words contain the subword CAB?

d. How many four-letter words begin with C or end in two vowels?

Solution

Note: As per given conditions, a word from this alphabet may have repeated letters. This is taken into account in what follows:

The alphabet has 6 letters, 2 vowels

a. How many four-letter words contain the subword ACE?

Number of four-letter words beginning with ACE = 6

Number of four letter words ending with ACE =6

So the number of four-letter words containing the subword ACE = 12

b. How many four-letter words don’t begin with F or don’t end in E?

The number of four letter words beginning with F and end in E= 6x6 =36

Total number of four-letter words using this alphabet = 6⁴ = 1296

So the number of four-letter words not beginning with F or don’t end in E = 1260

c. How many ve-letter words contain the subword CAB?

CAB (as a block) , may figure (i) in the beginning of the five letter word --in 36 words

                                               (ii) in the middle of the five letter word---in 36 words

                                               (iii) at the end of the five-letter word--in 36 words

Total number required =108

d. How many four-letter words begin with C or end in two vowels?

Number of four letter words beginning with C = 6³ =216

Number of words ending in two vowels = 6x6x2x2= 144

Number of wrods beginning with C and ending in two vowels = 6x2x2= 24.

So the number of four letter words beginning with C OR ending in two vowels = 216+144-24= 336