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2 $begingroup$ Suppose that we consider English alphabets so we have 26 letters which 5 of them are vowels. I want to find all three letters sequences that contain at least one vowel. My first approach is $26^3-21^3=8315$ which is the number of all three lettres sequences minus the number of three letter sequences which do not contain vowels. Second approach: at least one vowel means one vowel or two vowels or three vowels so the answer is $(5cdot21^2)+(5^2cdot21)+5^3=2855$ . Why are these two answers different? combinatorics discrete-mathematics share | cite | improve this question asked 2 hours ago Student St...