Arrangements without Consecutive Consonants in "INTERFERENCE"

Introduction
The task is to find the number of arrangements that can be made from the word "INTERFERENCE" such that no two consonants are together. To solve this problem, we will break it down into smaller steps and apply the principles of combinatorics.

Step 1: Identify Consonants and Vowels
First, we need to identify the consonants and vowels in the given word "INTERFERENCE". The consonants in this word are N, T, R, F, R, N, and C, while the vowels are I, E, E, E.

Step 2: Create a Template
To visualize the problem, let's create a template with placeholders for the consonants and vowels, where "C" represents a consonant and "V" represents a vowel:

C V C V C V C V C V C V C

Step 3: Place the Vowels
Since there are four vowels (I, E, E, E), we can arrange them in 4! (4 factorial) ways, which is equal to 24.

Step 4: Place the Consonants
Now we need to place the consonants in such a way that no two consonants are together. To achieve this, we can use the concept of "stars and bars" or "sticks and stones."

Step 4.1: Create Spaces for Consonants
We have 8 spaces available between the vowels in the template to place the consonants. Let's represent these spaces with "_":

C V C V C V C V C V C V C
_ _ _ _ _ _ _ _

Step 4.2: Distribute the Consonants
Now we need to distribute the 7 consonants (N, T, R, F, R, N, C) into these 8 spaces. We can think of this as placing the consonants in the spaces and treating the remaining spaces as "dividers" between the consonants.

Step 4.3: Applying Stars and Bars Principle
Using the stars and bars principle, we can calculate the number of ways to distribute the consonants. The formula is (n + k - 1)C(k - 1), where n is the number of objects to distribute (7 consonants) and k is the number of spaces (8 spaces).

Using the formula, we have (7 + 8 - 1)C(8 - 1) = 14C7 = 3432.

Step 5: Multiply the Possibilities
To find the total number of arrangements, we multiply the number of vowel arrangements (24) by the number of consonant arrangements (3432):

Total arrangements = 24 * 3432 = 82368

Conclusion
Therefore, there are 82,368 possible arrangements of the word "INTERFERENCE" where no two consonants are together. By breaking down the problem into smaller steps and applying combinatorial principles, we were able to find the solution.