Question

In how many ways can be the letter of the word ‘STRANGE’ be arranged so that
(a) The vowel may appear in the odd places
(b) The vowels are never separated
(c) The vowels never come together

Answer 1

Hint:There are 7 letters in the word ‘STRANGE’ in which two are vowels. There are four odd places. So, we have to find the total number of ways for these four places. The other 5 letters may be placed in the five places in 5! ways. By this we get the total number of required arrangements. If vowels are never separated, in this case consider all the vowels a single letter. If all the vowels never come together in this case, we subtract all vowels together from the total number of arrangements.

Complete step-by-step answer:
For finding number of ways of placing n things in r objects, we use ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Also, for arranging n objects without restrictions, we have $n!$ ways.
(a) There are 7 letters in the word ‘STRANGE’ amongst which 2 are vowels and there are 4 odd places (1, 3, 5, 7) where these two vowels are to be placed together.
Total number of ways can be expressed by replacing n = 4 and r = 2,
$\begin{align}
  & {}^{4}{{P}_{2}}=\dfrac{4!}{\left( 4-2 \right)!} \\
 & =\dfrac{4!}{2!} \\
 & =\dfrac{4\times 3\times 2!}{2!} \\
 & =4\times 3=12 \\
\end{align}$
And corresponding to these 12 ways the other 5 letters may be placed in 5! ways $=5\times 4\times 3\times 2\times 1=120$.
Therefore, the total number of required arrangements = $12\times 120=1440$ ways.

(b) Now, vowels are not to be separated. So, we consider all vowels as a single letter.
There are six letters S, T, R, N, G, (AE) , so they can arrange themselves in 6! ways and two vowels can arrange themselves in 2! ways.
Total number of required arrangements \[=6!\times 2!=6\times 5\times 4\times 3\times 2\times 1\times 2\times 1=1440\] ways.

(c) Now, for the number of arrangements when all vowels never come together, we subtract the total arrangement where all vowels have occurred together from the total number of arrangements.
The total number of arrangements = 7! = 5040 ways
And the number of arrangements in which the vowels do not come together $=7!-6!2!$
number of arrangements in which the vowels do not come together $=5040 -1440 = 3600$ ways.
Note: Knowledge of permutation and arrangement of objects under some restriction and without restriction is must for solving this problem. Students must be careful while making the possible cases. They must consider all the permutations in word like in part (b), a common mistake is done by not considering the arrangements of vowels.