Problem 35 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is

Step-by-Step Solution

Answer

300 arrangements where vowels do not come together.

Step 1: Calculate the Total Number of Arrangements

The word 'LETTER' consists of 6 letters where the letter 'T' is repeated twice. Using the formula for permutations of a multiset, the total number of arrangements is given by \( \frac{6!}{2!} \). This calculates as: \[ \frac{720}{2} = 360 \] Therefore, there are 360 possible arrangements of the letters.

Step 2: Treat Vowels as a Block

To ensure the vowels do not come together, we first calculate arrangements where they do come together. Consider the vowels 'EE' as a single entity or block, thus reducing the total to the following 'block' and consonants: L, T, T, R, (EE). This gives us 5 entities to arrange.

Step 3: Arrange the Block and Other Letters

Calculate the arrangements of these 5 entities, where 'T' is repeated. The arrangement will be \( \frac{5!}{2!} \). This works out to: \[ \frac{120}{2} = 60 \] Therefore, there are 60 arrangements where the 'EE' block is treated as one letter.

Step 4: Use Complementary Counting

To find arrangements where vowels do not come together, subtract the number of arrangements where vowels are together from the total arrangements calculated in Step 1. This gives: \( 360 - 60 = 300 \). Therefore, there are 300 arrangements where vowels do not come together.

Key Concepts

Factorial Calculations

Factorial calculations are fundamental in permutation and combination problems. Here, a factorial of a number is obtained by multiplying all positive integers up to that number. For example, the factorial of 6, denoted as \(6!\), is calculated as \(6 \times 5 \times 4 \times 3 \times 2 \times 1\), which equals 720.
In the context of permutations, factorials help determine how many ways a set of items can be arranged. When items include repetitions, as in the word "LETTER" with two 'T's, adjustments are made using factorals of the number of repeated items. Thus, the formula for permutations of a multiset becomes \( \frac{n!}{r_1!r_2! \ldots r_k!} \) where \(n\) is the total number of items, and each \(r_k!\) is the factorial of the count of each repeated item.
This approach reduces over-counting of permutations that appear identical due to repeated elements, as shown when calculating \( \frac{6!}{2!} \) for the word "LETTER", resulting in 360 possible arrangements.

Multisets

The concept of multisets involves the arrangement of items where some items may be identical. Unlike regular sets, the order of items matters in permutations of multisets.
In our exercise, dealing with the repeated 'T's in "LETTER" forms a multiset. While the distinct letters form a set, the repeated 'T's introduce a need for a permutation of a multiset. This requires acknowledging and adjusting for duplicates using the divisor in the factorial permutation calculation.

• This is effective when you need arrangements of letters like "L, E, T, T, E, R", where some letters appear more than once.
• Calculations for multisets ensure the distinct permutations are correctly counted.

This understanding helps manage complex arrangements in problems where not all items are unique, and establishes why factorial divisions (like \( \frac{6!}{2!} \)) are necessary to find the correct number of arrangements.

Complementary Counting

Complementary counting is a strategic method to solve counting problems, especially when direct counting is complex. Instead of counting the desired outcomes, you count the total outcomes first, then subtract the unwanted ones.
In our exercise, the goal was to arrange letters of "LETTER" such that vowels do not come together. Calculating directly for non-adjacent vowels can be cumbersome, so instead:

• Calculate all possible arrangements (360) first.
• Find arrangements where vowels ('EE') come together using a block method, yielding 60 arrangements.
• Subtract these from the total to find arrangements where vowels are separate, resulting in 300.

This complementary approach simplifies complex constraints by calculating their inverse, providing an efficient, accurate solution to permutation problems with intricate restrictions.