Permutations & CombinationsmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

4-Letter Words from INCONSEQUENTIAL = 3600 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The number of 44-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
A26702670
B28402840
C29202920
D36003600correct
Solution
Step 1: INCONSEQUENTIAL == I,N,C,O,N,S,E,Q,U,E,N,T,I,A,L. Distinct vowels {I,E,O,U,A}\{I,E,O,U,A\}: 55. Distinct consonants {N,C,S,Q,T,L}\{N,C,S,Q,T,L\}: 66. Step 2: (52)=5⋅42=10,(62)=6⋅52=15.\dbinom{5}{2}=\dfrac{5\cdot4}{2}=10,\qquad\dbinom{6}{2}=\dfrac{6\cdot5}{2}=15. Step 3: The 4 chosen letters are distinct ∴ 4!=244!=24 arrangements.
∴(52)(62)⋅4!=10⋅15⋅24=3600.\therefore\dbinom{5}{2}\dbinom{6}{2}\cdot4!=10\cdot15\cdot24=3600.
 Correct answer: (4)
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