From the letters of the word

Solution

ATQ,

Letters: vowels = A, A (2 identical); consonants = S, N, P, C, T (5 distinct). Arrange the 5 distinct consonants in a row. Number of ways = 5! = 120. This creates 6 gaps where vowels could be placed: _ C _ C _ C _ C _ C _ Gaps are: before the first consonant, 4 gaps between, and after the last consonant. Conditions. First letter must be a consonant ⇒ we cannot place a vowel in the very first gap (the gap before the first consonant), otherwise the word would start with a vowel. No two vowels adjacent ⇒ at most one vowel per gap. So the 2 vowels (both A’s) must be placed in 2 distinct gaps chosen from the remaining 5 allowed gaps (the 4 internal gaps between consonants and the gap after the last consonant). Number of ways to choose gaps for the two identical A’s = C(5, 2) = 10. Total arrangements = (arrangements of consonants) × (placements of vowels) = 5! × C(5, 2) = 120 × 10 = 1200.