Problem 25 - Algebra for JEE ADVANCED

Question

The number of words of four letters containing equal number of vowels and consonants, where repetition is allowed, is a. \(105^{2}\) b. \(210 \times 243\) c. \(105 \times 243\) d. \(150 \times 21^{2}\)

Step-by-Step Solution

Answer

The answer is 105^2.

Step 1: Identify the vowels and consonants

In the English alphabet, we have 5 vowels: a, e, i, o, and u. This means there are 26 - 5 = 21 consonants.

Step 2: Determine possible letter combinations

Since the word must have two vowels and two consonants, we need to calculate the number of 2-letter combinations for vowels and consonants separately, considering repetition is allowed.

Step 3: Calculate combinations for vowels

With repetition allowed, each of the two positions for vowels can be filled by any of the 5 vowels. Therefore, there are 5 choices for the first vowel and 5 choices for the second vowel: \[5 \times 5 = 25\] possible vowel combinations.

Step 4: Calculate combinations for consonants

Similarly, each of the two positions for consonants can be filled by any of the 21 consonants. Thus, there are 21 choices for the first consonant and 21 choices for the second consonant: \[21 \times 21 = 441\] possible consonant combinations.

Step 5: Calculate total number of words

To find the total number of 4-letter words with 2 vowels and 2 consonants, multiply the combinations for vowels by the combinations for consonants: \[ 25 \times 441 = 11025\].

Key Concepts

Combinatorics

Combinatorics is a fascinating branch of mathematics that deals with counting, arrangement, and combination of elements within a given set. It is crucial when addressing problems like how many different ways the letters of a word can be arranged. In our exercise, we're using combinatorics to count the number of possible four-letter words composed of equal numbers of vowels and consonants. To solve this, we first identify how many vowels (5) and consonants (21) we can use from the alphabet. By applying combinatorial principles, specifically involving combinations and repetitions, we determine the potential arrangements of these letters. Using combinatorics, we calculate possibilities considering that each letter position can be filled independently, allowing us to multiply the choices for vowels and consonants together for the total number of combinations.

Vowels and Consonants

When forming words using the English alphabet, distinguishing between vowels and consonants is fundamental. Vowels include the letters a, e, i, o, and u, while the remaining 21 letters are consonants. This distinction plays a key role in many word formation problems, as seen in the exercise, where the task is to create a four-letter word with exactly 2 vowels and 2 consonants. Understanding the balance between vowels and consonants helps in computing the number of valid combinations. In this exercise, since we need two of each, it requires calculating distinct sets of combinations separately for vowels and consonants, and then combining these results to find the total number of valid words.

Repetition in Combinations

Repetition in combinations allows each selected element to be used more than once when forming combinations, which significantly impacts the total number of arrangements. In this exercise, the letters can be repeated, meaning each vowel and consonant can appear more than once in the formation of the word.For vowels, with repetition, each of the two positions can be filled with any of the 5 vowels, leading to \[5 \times 5 = 25\]combinations. Similarly, each consonant position has 21 options, leading to \[21 \times 21 = 441\]combinations when considering repetition. This concept of repetition allows for a richer variety of combinations than if each letter could only be used once.

Word Formation

Word formation involves putting together letters in specific patterns based on given rules. In this example, the rule is to create a word of four letters comprising two vowels and two consonants, with letters allowed to repeat. To achieve this, we calculate the different ways to select and arrange vowels and consonants. The solutions involve combining the separate combinations calculated for vowels and consonants:

• Vowel combinations: 5 vowels repeated in 2 positions, yielding 25 combinations.
• Consonant combinations: 21 consonants repeated in 2 positions, yielding 441 combinations.

Finally, by multiplying these combinations, we find the total number of possible words. Through understanding word formation in these combinatory terms, students can learn how straightforward mathematical concepts aid in constructing various configurations.