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?
 A: 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.
A: Your calculation has a lot of overcounting, whenever there is more than one vowel present.
If you really want to avoid (or, say, cross-check) the "negative space" method of simply excluding options with no vowels, you could perhaps sum through the possibilities of where the first vowel is:


*

*Vowel in first place: $5\cdot 26^7 $ ways

*First vowel in second place: $21\cdot 5\cdot 26^6 $ ways

*First vowel in third place: $21^2\cdot 5\cdot 26^5 $ ways

*etc.


Total
$$5\cdot 26^7 + 21\cdot 5\cdot 26^6 + 21^2\cdot 5\cdot 26^5 + 21^3\cdot 5\cdot 26^4 \\
 \quad\quad+ 21^4\cdot 5\cdot 26^3 + 21^5\cdot 5\cdot 26^2 + 21^6\cdot 5\cdot 26 + 21^7\cdot 5 \\
= 5 \sum_{k=0}^7 21^k \,26^{7-k}$$
Not easy.

Another method is to calculate options for an exact number of vowels. This can be calculated by setting the pattern in one step, eg for three vowels:
$$BBABBAAB$$
which is ${8 \choose 3}$, then multiplying by the options for consonants and vowels respectively, so 
$${8 \choose 3}5^3\,21^5$$
for exactly three vowels. Then add up all options of interest (or, if simpler, add up the non-qualifying options and subtract from total).
