A word of at least `5` letters is made at random from `3` vowels and `3` constants, all the letters being different. The probability that no consonant falls between any two vowels in the word is

Options

A. `(9)/(20)`
B. `(9)/(10)`
C. `(7)/(10)`
D. `(11)/(20)`

Explanation

To solve the problem, we need to determine the probability that no consonant falls between any two vowels in a randomly formed word of at least 5 letters from 3 vowels and 3 consonants.

Let's break down the solution step by step:

### Step 1: Identify the total number of letters
We have 3 vowels (let's denote them as V1, V2, V3) and 3 consonants (C1, C2, C3). Therefore, we have a total of 6 letters.

### Step 2: Calculate the total number of 5-letter words
We can form words of 5 letters using different combinations of vowels and consonants. The possible combinations are:
1. 3 vowels and 2 consonants
2. 2 vowels and 3 consonants

#### Case 1: 3 Vowels and 2 Consonants
- The arrangement of 3 vowels can be done in \(3!\) ways.
- The arrangement of 2 consonants can be done in \(3C2\) ways (choosing 2 consonants from 3) and arranging them in \(2!\) ways.

So, the total arrangements for this case is:
\[ 3! \times \left(3C2 \times 2!\right) = 6 \times (3 \times 2) = 36 \]

#### Case 2: 2 Vowels and 3 Consonants
- The arrangement of 2 vowels can be done in \(3C2\) ways (choosing 2 vowels from 3) and arranging them in \(2!\) ways.
- The arrangement of 3 consonants can be done in \(3!\) ways.

So, the total arrangements for this case is:
\[ 3C2 \times 2! \times 3! = 3 \times 2 \times 6 = 36 \]

### Total arrangements for 5-letter words
Adding both cases together:
\[ 36 + 36 = 72 \]

### Step 3: Calculate the total arrangements where no consonant falls between vowels
To satisfy the condition that no consonant falls between any two vowels, we can arrange the vowels first and then place the consonants around them.

#### Arranging 3 Vowels
The arrangement of 3 vowels can be done in \(3!\) ways.

#### Arranging 2 Consonants
Once the vowels are arranged, we can place the consonants in the gaps around the vowels. For 3 vowels, there are 4 gaps (before the first vowel, between the vowels, and after the last vowel). We need to choose 2 out of these 4 gaps to place the consonants.

The number of ways to choose 2 gaps from 4 is \(4C2\), and the arrangement of the 2 consonants can be done in \(2!\) ways.

So, the total arrangements for this case is:
\[ 3! \times \left(4C2 \times 2!\right) = 6 \times (6 \times 2) = 72 \]

### Step 4: Calculate the total possible arrangements of 6 letters
The total number of arrangements of 6 letters (3 vowels + 3 consonants) is:
\[ 6! = 720 \]

### Step 5: Calculate the probability
The probability that no consonant falls between any two vowels is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{72}{720} = \frac{1}{10} \]

### Final Answer
The probability that no consonant falls between any two vowels in a randomly formed word of at least 5 letters is \(\frac{1}{10}\).

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