Question

Anagrams are made by using the letter of the word ‘HINDUSTAN’. In how many
of these anagrams all the vowels come together?

Answer 1

Hint: In English alphabet, there are five numbers of vowels. They are A, E, I, O and U. Rest
of the $21$ letters in the alphabet is known as ‘consonants’. Formula for permutation
is$P\left( {n,r} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. Formula for combination is,
$C\left( {n,r} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
Complete step by step answer:
Given word is ‘HINDUSTAN’.
Step I:
Total number of letters in the given word $ = 9$.
Number of vowels in the word are I, U, and A, that is $ = 3$
Numbers of consonants in the word are H, N, D, S, T, and N, that is $ = 6$
With internal arrangement, consider the vowels to be one entity with $3!$.
Hence, the total number of entities $ = 7$.
Step II:
Using the formula of permutation and combination,
Therefore, total number of arrangements $ = \dfrac{{7!}}{{2!}} \times 3!$
We are dividing 2! because the letter N in consonants is repeated twice.
Further solving,
$\begin{array}{c}\dfrac{{7!}}{{2!}} \times 3! = \dfrac{{7 \times 6 \times 5 \times 4 \times
3 \times 2 \times 1}}{{2!}} \times 3 \times 2!\\ = \dfrac{{15120 \times 2!}}{{2!}}\\ =
15120\end{array}$
Hence, vowels come together in $15120$ ways.
Note:
In step I, find the total numbers of vowels and the consonants in the given word. There are
five vowel quantities in the English alphabet. They are A, E, I, O, U. The 21 letters remaining in the alphabet are identified as consonants. In step II, apply the formulas of permutation and
combination. Permutations and combinations, the various ways in which objects from a set,
may be selected generally without replacement, to form subsets. This selection of subsets is
called a permutation when the order of selection is a factor, a combination when order is not a
factor. Obtain the total number of vowels.