Maths Q7 – JEE-MAIN 2026

Home Maths Permutations and CombinationsQuestion #323

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is:

Clue 1 of 3: Categorization

First, identify all unique letters in the given word and classify them into vowels and consonants.

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Step 1: Identify Unique Vowels and Consonants✦ Active

The unique letters in the word INCONSEQUENTIAL are I, N, C, O, S, E, Q, U, T, A, L. We categorize them into vowels and consonants:

Vowels (V): A, E, I, O, U (5 unique vowels)

Consonants (C): C, L, N, Q, S, T (6 unique consonants)

Step 2: Select Vowels and Consonants○ Expand

We need to form a 4-letter word with two vowels and two consonants. The number of ways to select 2 vowels from 5 unique vowels is $^{5} C_{2}$ , and the number of ways to select 2 consonants from 6 unique consonants is $^{6} C_{2}$ .

Number of ways to choose 2 vowels =^{5} C_{2} = \frac{5!}{2! 3!} = \frac{5 \times 4}{2} = 10

Number of ways to choose 2 consonants =^{6} C_{2} = \frac{6!}{2! 4!} = \frac{6 \times 5}{2} = 15

Total ways to choose 4 letters (2V + 2C) = 10 \times 15 = 150

Step 3: Arrange the Selected Letters○ Expand

Once 4 letters are chosen, they can be arranged in $4!$ ways to form a word. We multiply the number of ways to choose the letters by the number of ways to arrange them.

Number of ways to arrange 4 letters = 4! = 4 \times 3 \times 2 \times 1 = 24

Total number of 4-letter words = 150 \times 24 = 3600

💡 Teacher's Secret Hint

Ensure no letter repetition is considered, which is handled by using combinations and permutations of unique letters.

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Maths Question 7 – JEE-MAIN 2026

Clue 1 of 3: Categorization

🧠 Full Solution Path