Question

In how many ways can the letters of the word 'STRANGE' be arranged so that
(i) the vowels come together?
(ii) the vowels never come together? and
(iii) the vowels occupy only the odd places?

Answer 1

There are 7 letters in the word. STRANGE, including 2 vowels (A, E) and 5 consonants (S, T, R, N, G).
(i) Considering 2 vowels as one letter, we have 6 letters which can be arranged in $6_{P_6}$ = 6! ways A, E can be put together in 2! ways.
Hence, required number of words
= $6!\times 2!$
= $6\times 5\times 4\times 3\times 2\times 1\times 2$
= $720\times 2$
= 1440.
(ii) The total number of words formed by.: using all the letters of the words 'STRANGE'is $7_{P_7}=7!$
= $7\times 6\times 5\times 4\times 3\times 2\times 1$
= 5040.
So, the total number of words in which vowels are never together
= Total number of words - number of words in which vowels are always together
= 5040-1440
=3600
(iii) There are 7 letters in the word 'STRANGE', out of these letters 'A' and `E' are the vowels.
There are 4 odd places in the word 'STRANGE'. The two vowels can be arrangd in $4_{P_2}$ ways.
The remaining 5 consonants can be arranged among themselves in $5_{P_5}$ ways.
The total number of arrangements.
= $4_{P_2}\times 5_{P_5}$
= $\frac{4!}{2!}\times 5!$
= 1440