How many strings of $8$ English letters are there (repetition allowed)? a) at least one vowel b) start with $x$ and at least one vowel c) start and end with $x$ and at least one vowel I can solve them easily by considering $total-no$ $vowel$. So, a) $26^8 -21^8$ b) $26^7 -21^7$ c) $26^6 -21^6$ But, can I try a) choose $1$ vowel out of $5$ in $5$ ways and put in one of $8$ places, and then the remaining places $26^7$, which gives $5\cdot 8 \cdot 26^7$? What should be the logic thinking this way?

An alternative to Joffan's solution is to count up all the ways there could be exactly $k$ vowels (as suggested by André Nicolas). We then get

$$ N = \sum_{k=1}^8 \binom{8}{k} 5^k 21^{8-k} $$

All the methods yield $N = 171004205215$, confirming the expression $26^8-21^8$ you originally derived.