Question

a)     How many four-letter words can be formed from the letters of the word LAUNDRY if each letter can only be used one time in a word? Y is NOT considered a vowel in this word.

b)    How many contain the letter Y?

c)     How many contain both vowels?

d)    How many of them contain exactly three consonants?

e)    How many of them begin and end in a consonant?

f)      How many begin with N and end in a vowel?

g)     How many begin with N and also contain Y?

h)    How many of them contain both N and Y?

Answer 1

The length of the word "LAUNDRY" is 7 with 2 vowels A and U, the remaining 5 are constants.

a) Number of four-letter words = (ways of choosing 4 words from 7)*(ways of arranging these 4 letters)

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b) Here letter Y is fixed to be taken, thus we have to choose 3 letters from the remaining 6 letters. hence the number of words containing Y = (ways of choosing 3 words from 6)*(ways of arranging these 4 letters)

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c) There are 2 vowels, thus we have to choose 2 letters from the remaining 5 letters. hence the number of words containing both the vowels= (ways of choosing 2 words from 5)*(ways of arranging these 4 letters)

d) Here we have to choose three consonants out of 5 and 1 vowel from 2. Thus, the number of words with exactly 3 consonants and a vowel is: (ways of choosing 3 consonants and a vowel) * (ways of arranging these 4 letters)

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e) Here we're fixing that the first and last words have to be consonants, thus first we'll choose to consonants in these places and the remaining two places don't have any restriction. Thus the number of words which begin and end in a consonant is

= (ways of choosing 2 consonants for end a begin) * (ways of choosing two letters from remaining from 5 letters)*

(ways of arranging 2 letters at the end the beginning) * (ways of arranging in between 2 letters)

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f) Here N is fixed to be at the beginning and then we choose a vowel at the end while the remaining two letters are selected from five remaining letters. Thus, the number of words which begin with and end in a vowel

= (selecting a vowel a the end) * (selecting two remaining letters) * (ways of arranging in between 2 letters)

=

g) Here N is fixed to be at the beginning and then Y is also selected then the remaining two letters are selected from five remaining letters. thus, the number of words begin with N and also contain Y

= (selecting two remaining letters) * (ways of arranging 3 letters)

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h) Here the letters N and Y are fixed to be there thus selecting two remaining letters. hence number of words contain both N and Y = (selecting two remaining letters) * (ways of arranging these 4 letters)

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