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Question: How many anagrams can be made by using the letters of the word HINDUSTAN? How many of these anagrams...

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?

Explanation

Solution

To find: We have to find the total anagrams which can be made by using the 99 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 99 letters are used. Out of these 99 letters, 33 letters, are vowels and 66 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) =9!2! = \dfrac{{9!}}{{2!}} because 22 consonants are alike therefore we have to divide the outcomes by 9!9!
Hence total anagrams possible =9!2! = \dfrac{{9!}}{{2!}}
=9×8×7×6×5×4×3×2×12×1 =181,440  = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \\\ = 181,440 \\\
Hence 181,440181,440 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 33 vowels and 66 consonants, in which 22 consonants are alike.
So the first place can be filled in 33 ways and the last place can be filled in 22 ways.
\therefore the remaining places can be filled in =7!2! = \dfrac{{7!}}{{2!}} ways.
Hence the required anagrams =3×2×7!2! = 3 \times 2 \times \dfrac{{7!}}{{2!}}
=3×2×7×6×5×4×3×2×12×1 =15120  = \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!2! in both the cases. We need to be careful while doing calculations of factorials.