Question

How many permutations can be formed by the letters of the word, 'VOWELS', when
(i) there is no restriction on letters?
(ii) each word begins with E?
(iii) each word begins with O and ends with L?
(iv) all vowels come together?
(v) all consonants come together?

Answer 1

There are 6 letters in the word 'VOWELS'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal to ${}^6{P}_6=6!=6\times 5\times 4\times 3\times 2\times 1=720$

(ii) If we fix up E in the beginning then the remaining 5 letters can be arranged in ${}^5{P}_5=5!=5\times 4\times 3\times 2\times 1=120$ ways

(iii) If we fix up O in the beginning and L at end, the remaining 4 letters can be deranged in ${}^4{P}_4=4!=4\times 3\times 2\times 1=24$.

(iv) There are 2 vowels and 4 consonants in the word 'VOWELS'. Considering 2 vowels as one letter, we have letters which can be arranged in ${}^5{P}_5=5!$ ways,

O, E can be put together in 2! ways. Hence, required number of words = $5!\times 2!$

=$5\times 4\times 3\times 2\times 1\times 2\times 1=120\times 2=240$

(v) There are 2 vowels and 4 consonants the word 'VOWELS'. Considering 4 consonants as one letter, we have 3 letters which can be arranged in ${}^3{P}_3$ = 3! ways.

U, W, L, S can be put together in 4! ways, Hence, required number of words in which all consonants come together = $3!\times 4!=3\times 2\times 4\times 3\times 2=144$