find the number of seven letter "words" that start with a or do not contain a

Question

Find the number of seven letter "words" that start with a or do not contain a. For this problem, I'm not really sure what method I can use. I tried to subtract the "words" that start with a from 657800 (26 choose 7)

Answer 1

"Words" here should be taken to mean arrangements of letters, not actual dictionary words.

The first case is having an "a" at the start. Which means that the remaining six postions can contain any letter (including "a"). This gives $1\cdot 26^6 = 26^6$ possibilities.

The second case is not containing an "a" at all. Which means all seven positions must be filled with one of the remaining $25$ letters in the alphabet. That gives $25^7$ possibilities.

Note that the two cases are mutually exclusive.

Therefore the total number is $26^6 + 25^7 =6412431401$.

EDIT: Addressing your attempted answer, note that $\binom {26}7$ would only be relevant if you're choosing $7$ (unique, non-repeated) letters out of the alphabet to fill the $7$ positions without regard to order. This is not a relevant consideration here.

Answer 2

So, answer should be as follows:

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Here we have $7$ spaces. The total number of words that start with a are : $1 \cdot 26^6$ ($1$ is representing a here and $26$ representing rest of letters) And the words that don't contain a are: $25^7$ (all words other than a). See there is no intersection between two cases. Like you can't have a word that start with a and do not contain a. Right? So, answer will be $26^6+25^7$. Thanks, hope it helps.