Question

How many anagrams can be made by using the letters of the word HINDUSTAN? How many of these anagrams begin and end with a vowel?

Answer 1

Hint: To find: We have to find the total anagrams which can be made by using the $9$ letters of the word HINDUSTAN and also we have to find that how many of these anagrams begin and with a vowel.

Complete Step-by-step Solution
A word HINDUSTAN is given in which $9$ letters are used. Out of these $9$ letters, $3$ letters, are vowels and $6$ letters are consonants.
The word is given HINDUSTAN.
In this word two letters alike i.e. N.
The total anagrams can be made by using a given letters (H, I, N, D, U, S, T, A, N) $ = \dfrac{{9!}}{{2!}}$ because $2$ consonants are alike therefore we have to divide the outcomes by $9!$
Hence total anagrams possible $ = \dfrac{{9!}}{{2!}}$
$
   = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \\
   = 181,440 \\
 $
Hence $181,440$ anagrams can be made.
In this case, fix the first place and last place with a vowel and then find the remaining places with permutation concept and then multiply all the outcomes.
Now we will find the anagrams begin and end with a vowel.
We have $3$ vowels and $6$ consonants, in which $2$ consonants are alike.
So the first place can be filled in $3$ ways and the last place can be filled in $2$ ways.
$\therefore $ the remaining places can be filled in $ = \dfrac{{7!}}{{2!}}$ ways.
Hence the required anagrams $ = 3 \times 2 \times \dfrac{{7!}}{{2!}}$
$
   = \dfrac{{3 \times 2 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \\
   = 15120 \\
$

Note:
In word HINDUSTAN the letter ‘N’ comes two times therefore we divided the outcomes by $2!$ in both the cases. We need to be careful while doing calculations of factorials.