Question

Using the letters of the word HINDUSTAN, how many 9 letter words will have none of the vowels together?

Answer 1

75600

Answer 2

To solve this problem, we use the gap method. The word HINDUSTAN has 9 letters.

First, let's categorize the letters into vowels and consonants and identify any repetitions: The letters in HINDUSTAN are: H, I, N, D, U, S, T, A, N.

1. Identify Vowels (V) and Consonants (C):

   • Vowels: I, U, A (3 distinct vowels)
   • Consonants: H, N, D, S, T, N (6 consonants)

2. Identify Repeated Letters:

   • The letter 'N' appears twice among the consonants. All other letters are unique.

To ensure none of the vowels are together, we follow these steps:

Step 1: Arrange the Consonants First, arrange the 6 consonants: H, N, D, S, T, N. Since 'N' is repeated twice, the number of ways to arrange these consonants is given by the permutation formula for objects with repetitions: Number of ways to arrange consonants = $\frac{\text{(Total number of consonants)}!}{\text{(Frequency of repeated letter 1)}! \times \text{(Frequency of repeated letter 2)}! \times \dots}$ Number of ways to arrange consonants = $\frac{6!}{2!} = \frac{720}{2} = 360$ ways.

Step 2: Create Gaps for Vowels When we arrange 6 consonants, there are 7 possible positions (gaps) where the vowels can be placed so that no two vowels are adjacent. Let 'C' represent a consonant. The arrangement of consonants creates gaps as follows: _ C _ C _ C _ C _ C _ C _ There are 7 gaps (represented by _).

Step 3: Arrange the Vowels in the Gaps We have 3 distinct vowels (I, U, A) to place in 3 of the 7 available gaps. Since the vowels are distinct and their order matters, this is a permutation problem. Number of ways to choose and arrange 3 distinct vowels in 7 distinct gaps = $P(7, 3)$ or $^7P_3$. $^7P_3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$ ways.

Step 4: Calculate the Total Number of Words To find the total number of 9-letter words where none of the vowels are together, multiply the number of ways to arrange the consonants by the number of ways to place the vowels in the gaps. Total number of words = (Ways to arrange consonants) $\times$ (Ways to place vowels in gaps) Total number of words = $360 \times 210 = 75600$.