A problem in my textbook says "How many strings of six lowercase letters from the English alphabet contain the letter 'a'?"
A proposed solution was $26^{6} - 25^{6}$.
However, since the letter a can be placed in 6 different way, and I can arrange the rest of the alphabet in 5 remaining positions, shouldn't I be allowed to do $C(6,1) *25 * 24 * 23 * 22 *21$ ?
What am I miscounting in the second approach?
On
It seems that you are assuming that '$a$' appears once and every letter can be used exactly once, which is not assumed in the question.
The proposed solution first consider the number of possiblities without any constraint and then we subtract the number of possibilities where no '$a$' appears.
$C(6,1)$ means that you only count words containing one "a". What about words containing two "a"s? or three? or four? or five? or six?
Additionally, your decreasing product for the other letters means you are counting as if you can only use at most one of the non-"a" letters in such a word.
So, you do not count "llamas", "access", "aaaaaa", "ackack", "affect", "attack", "innate" et al. ... and those are actual words. You also doun't count "qzaqzq".