Permutations With Restrictions

Arranging letters under positional restrictions is handled by first placing the restricted letters, then the rest, a standard JEE Advanced permutation strategy. The word MONDAY has 6 distinct letters with two vowels O and A and four consonants M, N, D, Y. There are three even positions, the second, fourth and sixth. The two vowels must occupy even positions, so choose and arrange 2 of the 3 even slots for the vowels in P(3,2) = 3 × 2 = 6 ways. The remaining 4 positions, namely the three odd positions plus one leftover even position, are filled by the 4 consonants in 4! = 24 ways. By the multiplication principle, the total is 6 × 24 = 144. Option 72 forgets one factor in arranging vowels. Option 120 = 5! ignores the restriction. Option 36 omits the consonant arrangements. Hence the answer is 144. Plausibility check: the count must be far below the unrestricted 6! = 720, and 144 is exactly one fifth of it, consistent with constraining two vowels to a subset of positions. Positional-restriction problems are handled most reliably by first placing the constrained letters into their allowed slots and only afterwards filling the remaining positions, which prevents double counting. The deeper principle is that the multiplication rule applies whenever independent stages are performed in sequence, so identifying genuinely independent choices is the key judgement these arrangement questions are testing.