Q32 / 75 · JEE Main 2026 (06 April Shift 1) · Mathematics
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:
- A$2670$
- B$2840$
- C$2920$
- Correct answerD$3600$
Solution
The given word is INCONSEQUENTIAL.
The vowels present in the word are I, O, E, U, A. There are $5$ distinct vowels.
The consonants present in the word are N, C, S, Q, T, L. There are $6$ distinct consonants.
We need to form $4$-letter words consisting of $2$ vowels and $2$ consonants without repeating any letter. This means all $4$ letters in the formed word must be distinct.
Number of ways to select $2$ distinct vowels from $5$ distinct vowels = $^{5}C_{2} = \dfrac{5 \times 4}{2} = 10$
Number of ways to select $2$ distinct consonants from $6$ distinct consonants = $^{6}C_{2} = \dfrac{6 \times 5}{2} = 15$
Number of ways to arrange the $4$ selected letters = $4! = 24$
Total number of $4$-letter words that can be formed = $10 \times 15 \times 24 = 3600$
Answer: $3600$
The vowels present in the word are I, O, E, U, A. There are $5$ distinct vowels.
The consonants present in the word are N, C, S, Q, T, L. There are $6$ distinct consonants.
We need to form $4$-letter words consisting of $2$ vowels and $2$ consonants without repeating any letter. This means all $4$ letters in the formed word must be distinct.
Number of ways to select $2$ distinct vowels from $5$ distinct vowels = $^{5}C_{2} = \dfrac{5 \times 4}{2} = 10$
Number of ways to select $2$ distinct consonants from $6$ distinct consonants = $^{6}C_{2} = \dfrac{6 \times 5}{2} = 15$
Number of ways to arrange the $4$ selected letters = $4! = 24$
Total number of $4$-letter words that can be formed = $10 \times 15 \times 24 = 3600$
Answer: $3600$
Hence, the correct answer is option: 'D'.